Exam 2 Practice
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Mole, Percent Composition, Percent Abundance
Which one of the samples contains the most particles?
- 1 mol CO2(g)
- 1 mol UF6(g)
- 1 mol CH3COCH3(l)
- 1 mol He(g)
- all contain the same number of particles
Solution
Answer: E
Since there is 1 mole of each sample, each sample contains 6.022 × 1023 particles.
Concept: number of entities; moles
Which one of the samples has the largest mass?
- 1 mol CO2(g)
- 1 mol UF6(g)
- 1 mol CH3COCH3(l)
- 1 mol He(g)
- all contain the same number of particles
Solution
Answer: B
Since each sample contains the same number of particles, identify the particle with the largest mass. UF6 has the largest mass and therefore, the 1 mole sample has the largest mass.
Concept: number of entities; moles
Which one of the samples contains the most atoms?
- 1 mol CO2(g)
- 1 mol UF6(g)
- 1 mol CH3COCH3(l)
- 1 mol He(g)
- all contain the same number of particles
Solution
Answer: C
\[\begin{align*} N(X) &= n(\mathrm{CO_2})~ r(X\mathrm{,CO_2})~ N_{\mathrm{A}} \\[1.5ex] &= 1~\mathrm{mol~CO_2} \left ( \dfrac{3~\mathrm{mol}~X}{\mathrm{mol~CO_2}} \right ) \left ( \dfrac{N_0}{\mathrm{mol}} \right )\\[1.5ex] &= 4~\!N_0\\[2ex] \mathbf{B}: N(X) &= n(\mathrm{UF_6}) ~ r(X\mathrm{,UF_6}) ~ N_{\mathrm{A}} \\[1.5ex] &= 1~\mathrm{mol~UF_6} \left ( \dfrac{7~\mathrm{mol}~X}{\mathrm{mol~UF_6}} \right ) \left ( \dfrac{N_0}{\mathrm{mol}} \right )\\[1.5ex] &= 7~\!N_0\\[2ex] \mathbf{C}: N(X) &= n(\mathrm{CH_3COCH_3}) ~ r(X\mathrm{,CH_3COCH_3}) ~ N_{\mathrm{A}} \\[1.5ex] &= 1~\mathrm{mol~CH_3COCH_3} \left ( \dfrac{10~\mathrm{mol}~X}{\mathrm{mol~CH_3COCH_3}} \right ) \left ( \dfrac{N_0}{\mathrm{mol}} \right )\\[1.5ex] &= 10~\!N_0\\[2ex] \mathbf{D}: N(X) &= n(\mathrm{He}) ~ N_{\mathrm{A}} \\[1.5ex] &= 1~\mathrm{mol}~X \left ( \dfrac{N_0}{\mathrm{mol}} \right )\\[1.5ex] &= N_0 \end{align*}\]
Concept: number of entities; moles
What is the molar mass (in g mol–1 to four significant figures) of Al2(SO4)3 · 18 H2O?
Solution
Answer: 662.5 g mol–1
\[\begin{align*} M[\mathrm{Al_2(SO_4)_3 \boldsymbol{\cdot} 18~H_2O}] &= [ 2~A_{\mathrm{r}}^{\circ}(\mathrm{Al}) + 3~A_{\mathrm{r}}^{\circ}(\mathrm{S}) + 30~A_{\mathrm{r}}^{\circ}(\mathrm{O}) + 36~A_{\mathrm{r}}^{\circ}(\mathrm{H}) ]~ M_{\mathrm{u}} \\[1.5ex] &= [ 2 \left ( 26.98 \right ) + 3 \left ( 32.06 \right ) + 30 \left ( 16.00 \right ) + 36 \left ( 1.01 \right ) ] \left ( \dfrac{\mathrm{g}}{\mathrm{mol}} \right ) \\[1.5ex] &= 666.\bar{4}6~\mathrm{g~mol^{-1}} \\[1.5ex] &= 666.5~\mathrm{g~mol^{-1}} \end{align*}\]
Concept: molar mass
What is the molar mass (in g mmol–1 to three significant figures) of (NH4)2HPO4?
Solution
Answer: 0.132 g mmol–1
\[\begin{align*} M[\mathrm{(NH_4)_2HPO_4}] &= [ 2~A_{\mathrm{r}}^{\circ}(\mathrm{N}) + 9~A_{\mathrm{r}}^{\circ}(\mathrm{H}) + A_{\mathrm{r}}^{\circ}(\mathrm{P}) + 4~A_{\mathrm{r}}^{\circ}(\mathrm{O}) ]~ M_{\mathrm{u}} \\[1.5ex] &= [ 2 \left ( 14.01 \right ) + 9 \left ( 1.01 \right ) + \left ( 30.97 \right ) + 4 \left ( 16.00 \right ) ] \left ( \dfrac{\mathrm{g}}{\mathrm{mol}} \right ) \left ( \dfrac{\mathrm{mol}}{10^3~\mathrm{mmol}} \right ) \\[1.5ex] &= 0.13\bar{2}08~\mathrm{g~mmol^{-1}} = 0.132~\mathrm{g~mmol^{-1}} \end{align*}\]
Concept: molar mass
How many atoms are present in a 0.268 mol sample of CH3OH?
Solution
Answer: 9.68 × 1023
\[\begin{align*} N(X) &= n(\mathrm{CH_3OH})~ r(\mathrm{X,CH_3OH})~ N_{\mathrm{A}} \\[1.5ex] &= 0.268~\mathrm{mol~CH_3OH} \left ( \dfrac{6~\mathrm{mol}~X}{\mathrm{mol~CH_3OH}} \right ) \left ( \dfrac{6.022\times 10^{23}~X}{\mathrm{mol}~X} \right ) \\[1.5ex] &= 9.6\bar{8}33\times 10^{23} = 9.68\times 10^{23} \end{align*}\]
Concept: number of particles; moles
How many aluminum atoms are there in a 3.50 g sample of Al2O3?
Solution
Answer: 4.13 × 1022
\[\begin{align*} N(\mathrm{Al}) &= m(\mathrm{Al_2O_3}) ~ M(\mathrm{Al_2O_3})^{-1} ~ r(\mathrm{Al,Al_2O_3}) ~ N_{\mathrm{A}} \\[1.5ex] &= 3.50~\mathrm{g~Al_2O_3} \left ( \dfrac{\mathrm{mol~Al_2O_3}}{101.96~\mathrm{g}} \right ) \left ( \dfrac{2~\mathrm{mol~Al}}{\mathrm{mol~Al_2O_3}} \right ) \left ( \dfrac{6.022\times 10^{23}~\mathrm{Al}}{\mathrm{mol~Al}} \right ) \\[1.5ex] &= 4.1\bar{3}43\times 10^{22} \\[1.5ex] &= 4.13\times 10^{22} \end{align*}\]
Concept: number of particles; moles
A sample of a compound containing only carbon and oxygen decomposes and produces 24.5 g of carbon and 32.59 g of oxygen. What is the sample?
- CO
- CO2
- CO3
- C3O2
- C2O
Solution
Answer: A
Total mass:
\[\begin{align*} m(\mathrm{sample}) &= 24.5~\mathrm{g} + 32.59~\mathrm{g} = 57.09~\mathrm{g} \\ \end{align*}\]
Mass percent of C and O in sample:
\[\begin{align*} w(\mathrm{C})\% &= w(\mathrm{C}) \times 100~\% \\[1.5ex] &= \dfrac{m(\textrm{C})}{m(\mathrm{sample})} \times 100~\% \\[1.5ex] &= \dfrac{24.5~\mathrm{g}}{57.09~\mathrm{g}} \times 100~\% \\[1.5ex] &= 42.\bar{9}14~\% \\[3ex] w(\mathrm{O})\% &= w(\mathrm{O}) \times 100~\% \\[1.5ex] &= \dfrac{m(\textrm{O})}{m(\mathrm{sample})} \times 100~\% \\[1.5ex] &= \dfrac{32.59~\mathrm{g}}{57.09~\mathrm{g}} \times 100~\% \\[1.5ex] &= 57.0\bar{8}53~\% \end{align*}\]
Mass percent of C and O in CO:
\[\begin{align*} w(\mathrm{C})\% &= w(\mathrm{C}) \times 100~\% \\[1.5ex] &= \dfrac{A_{\textrm{r}}^{\circ}(\textrm{C})}{A_{\textrm{r}}^{\circ}(\textrm{CO})} \times 100~\% \\[1.5ex] &= \dfrac{12.01}{28.01} \times 100~\% \\[3ex] &= 42.8\bar{7}75~\% \\[3ex] w(\mathrm{O})\% &= w(\mathrm{O}) \times 100~\% \\[1.5ex] &= \dfrac{A_{\textrm{r}}^{\circ}(\textrm{O})}{A_{\textrm{r}}^{\circ}(\textrm{CO})} \times 100~\% \\[1.5ex] &= \dfrac{16.00}{28.01} \times 100~\% \\[1.5ex] &= 57.1\bar{2}24~\%\\[3ex] \end{align*}\]
The mass percent of C and O in the sample closely match the mass percent of C and O in CO. The other options give significantly different mass percent compositions.
Concept: Law of Definite Proportions; percent by mass; mole ratio
Guanidine, HNC(NH2)2, is a fertilizer. What is the percent by mass (to one decimal place) of nitrogen in the fertilizer?
Solution
Answer: 71.1 %
\[\begin{align*} w(\mathrm{N})\% &= w(\mathrm{N}) \times 100~\% \\[1.5ex] &= \Biggl\{\dfrac{3~m(\mathrm{N})}{m(\mathrm{HNC(NH_2)_2})} \Biggl\}\times 100~\% \\[1.5ex] &= \Biggl\{ \dfrac{3\!~A_{\mathrm{r}}^{\circ}(\mathrm{N})} { \left [ 5~A_{\mathrm{r}}^{\circ}(\mathrm{H}) \right ] + \left [ 3~A_{\mathrm{r}}^{\circ}(\mathrm{N}) \right ] + A_{\mathrm{r}}^{\circ}(\mathrm{C}) } \Biggl\} \times 100~\% \\[1.5ex] &= \Biggl\{ \dfrac{3 \left(14.01 \right)} { \left ( 5\times 1.01 \right ) + \left ( 3\times 14.01 \right ) + 12.01 } \Biggl\} \times 100~\% \\[1.5ex] &= \left ( \dfrac{42.03}{59.09} \right ) \times 100~\% \\[1.5ex] &= 71.\bar{1}28~\% \\[1.5ex] &= 71.1~\% \end{align*}\]
Concept: percent composition by mass
Determine the percent by mass (to one decimal place) of Mg in chlorophyll (C55H72MgN4O5), the green pigment in plant cells.
Solution
Answer: 2.7 %
\[\begin{align*} w(\mathrm{Mg})\% &= w(\mathrm{Mg}) \times 100~\% \\[1.5ex] &= \Biggl\{ \dfrac{m(\mathrm{Mg})}{m(\mathrm{chlorophyll})} \Biggl\} \times 100~\% \\[1.5ex] &= \Biggl\{ \dfrac{A_{\mathrm{r}}^{\circ}(\mathrm{Mg})} { \left [ 55\times A_{\mathrm{r}}^{\circ}(\mathrm{C}) \right ] + \left [ 72\times A_{\mathrm{r}}^{\circ}(\mathrm{H}) \right ] + A_{\mathrm{r}}^{\circ}(\mathrm{Mg}) + \left [ 4\times A_{\mathrm{r}}^{\circ}(\mathrm{N}) \right ] + \left [ 5\times A_{\mathrm{r}}^{\circ}(\mathrm{O}) \right ] } \Biggl\} \times 100~\% \\[1.5ex] &= \Biggl\{ \dfrac{ 24.31 } { \left ( 55\times 12.01 \right ) + \left ( 72\times 1.01 \right ) + 24.31 + \left ( 4\times 14.01 \right ) + \left ( 5\times 16.00 \right ) } \Biggl\} \times 100~\% \\[1.5ex] &= \Biggl\{ \dfrac{24.31}{893.62} \Biggl\} \times 100~\% \\[1.5ex] &= 2.\bar{7}20~\% \\[1.5ex] &= 2.7~\% \end{align*}\]
Concept: percent composition by mass
The mineral spodumene has the formula LiAlSi2O6 . What is the mass (in g to two decimal places) of lithium in a 438 g sample?
Solution
Answer: 16.3 g
\[\begin{align*} m(\mathrm{Li}) &= w(\mathrm{Li})~ m(\mathrm{sample}) \\[1.5ex] &= \left ( \dfrac{m(\mathrm{Li})}{m(\mathrm{LiAlSi_2O_6})} \right ) ~ m(\mathrm{sample}) \\[1.5ex] &= \left ( \dfrac{A_{\mathrm{r}}^{\circ}(\mathrm{Li})} { A_{\mathrm{r}}^{\circ}(\mathrm{Li}) + A_{\mathrm{r}}^{\circ}(\mathrm{Al}) + \left [ 2~A_{\mathrm{r}}^{\circ}(\mathrm{Si}) \right ] + \left [ 6~A_{\mathrm{r}}^{\circ}(\mathrm{O}) \right ] } \right ) ~ m(\mathrm{sample}) \\[1.5ex] &= \left ( \dfrac{6.94}{186.1}\right ) \left ( \dfrac{438~\mathrm{g}}{} \right ) \\[1.5ex] &= 16.\bar{3}33~\mathrm{g}\\[1.5ex] &= 16.3~\mathrm{g} \end{align*}\]
Concept: mass fraction
How many moles (in normalized scientific notation) of Cs are contained in 595 kg of Cs?
Solution
Answer: 4.48 × 103 mol
\[\begin{align*} n(\mathrm{Cs}) &= m(\mathrm{Cs}) ~ M(\mathrm{Cs})^{-1} \\[1.5ex] &= 595~\mathrm{kg} \left ( \dfrac{10^3~\mathrm{g}}{\mathrm{kg}} \right ) \left ( \dfrac{\mathrm{mol}}{132.91~\mathrm{g}} \right ) \\[1.5ex] &= 44\bar{7}6.7~\mathrm{mol} \\[1.5ex] &= 4.48\times 10^{3}~\mathrm{mol} \end{align*}\]
Concept: moles
Analysis of a sample of a covalent compound showed that it contained 14.4 % hydrogen and 85.6 % carbon by mass. What is the empirical formula for the compound?
Solution
Answer: CH2
Here I choose the mass of the sample to be exactly 100 g.
\[\begin{align*} w(\mathrm{H})\% &= w(\mathrm{H}) \times 100~\% \\[1.5ex] &= m(\mathrm{H})~ m(\mathrm{sample})^{-1}~ \times 100~\% \\[1.5ex] &= n(\mathrm{H}) ~ M(\mathrm{H})~ m(\mathrm{sample})^{-1}~ \times 100~\% \longrightarrow \\[1.5ex] n(\mathrm{H}) &= \left ( \dfrac{w\left(\mathrm{H}\right)\%}{100~\%} \right ) ~ m(\mathrm{sample}) ~ M(\mathrm{H})^{-1} \\[1.5ex] &= \left ( \dfrac{14.4~\%}{100~\%}\right ) \left ( \dfrac{100~\mathrm{g}}{} \right ) \left ( \dfrac{\mathrm{mol}}{\mathrm{1.01~g}} \right ) \\[1.5ex] &= 14.\bar{2}57~\mathrm{mol~H}\\[3ex] w(\mathrm{C})\% &= w(\mathrm{C}) \times 100~\% \\[1.5ex] &= m(\mathrm{C})~ m(\mathrm{sample})^{-1}~ \times 100~\% \\[1.5ex] &= n(\mathrm{C}) ~ M(\mathrm{C})~ m(\mathrm{sample})^{-1}~ \times 100~\% \longrightarrow \\[1.5ex] n(\mathrm{C}) &= w(\mathrm{C}) ~ m(\mathrm{sample}) ~ M(\mathrm{C})^{-1} \\[1.5ex] &= \left ( \dfrac{w\left(\mathrm{C}\right)\%}{100~\%} \right ) ~ m(\mathrm{sample}) ~ M(\mathrm{C})^{-1} \\[1.5ex] &= \left ( \dfrac{85.6~\%}{100~\%}\right ) \left ( \dfrac{100~\mathrm{g}}{} \right ) \left ( \dfrac{\mathrm{mol}}{\mathrm{12.01~g}} \right ) \\[1.5ex] &= 7.1\bar{2}7~\mathrm{mol~C}\\[3ex] r(\mathrm{H,C}) &= n(\mathrm{H})~ n(\mathrm{C})^{-1}\\[1.5ex] &= \dfrac{14.\bar{2}57~\mathrm{mol~H}}{7.1\bar{2}7~\mathrm{mol~C}}\\[1.5ex] &= \dfrac{2.00~\mathrm{mol~H}}{1~\mathrm{mol~C}} \\[2ex] &\mathrm{CH_2} \end{align*}\]
Concept: percent composition by mass; empirical formula
What is the mass (in g) of 2.6 × 1022 chlorine atoms?
Solution
Answer: 1.5 g
\[\begin{align*} m(\mathrm{Cl}) &= n(\mathrm{Cl}) ~ M(\mathrm{Cl})\\[1.5ex] &= N(\mathrm{Cl}) ~ N_{\mathrm{A}}^{-1} ~ M(\mathrm{Cl}) \\[1.5ex] &= 2.6\times 10^{22}~\mathrm{Cl} \left ( \dfrac{\mathrm{mol}}{6.022\times 10^{23}~\mathrm{Cl}} \right ) \left ( \dfrac{35.45~\mathrm{g}}{\mathrm{mol}} \right ) \\[1.5ex] &= 1.\bar{5}30~\mathrm{g} \\[1.5ex] &= 1.5~\mathrm{g} \end{align*}\]
Concept: moles
How many iron atoms are contained in 354 g of iron?
Solution
Answer: 3.82 × 1024
\[\begin{align*} N(\mathrm{Fe}) &= n(\mathrm{Fe}) ~ M(\mathrm{Fe}) \\[1.5ex] &= m(\mathrm{Fe}) ~ M(\mathrm{Fe})^{-1} ~ N_{\mathrm{A}} \\[1.5ex] &= 354~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{55.85~\mathrm{g}} \right ) \left ( \dfrac{6.022\times 10^{23}~\mathrm{Fe}}{\mathrm{mol}} \right ) \\[1.5ex] &= 3.8\bar{1}69 \times 10^{24}~\mathrm{Fe}\\[1.5ex] &= 3.82\times 10^{24}~\mathrm{Fe} \end{align*}\]
Concept: moles; number of particles
What is the mass (in ng) of 2.33 × 1020 oxygen atoms?
Solution
Answer: 6.19 × 106 ng
\[\begin{align*} m(\mathrm{O}) &= n(\mathrm{O}) ~ M(\mathrm{O}) \\[1.5ex] &= N(\mathrm{O}) ~ N_{\mathrm{A}}^{-1} ~ M(\mathrm{O}) \\[1.5ex] &= 2.33\times 10^{20}~\mathrm{O} \left ( \dfrac{\mathrm{mol}}{6.022\times 10^{23}~\mathrm{O}} \right ) \left ( \dfrac{16.00~\mathrm{g}}{\mathrm{mol}} \right ) \left ( \dfrac{10^9~\mathrm{ng}}{\mathrm{g}} \right ) \\[1.5ex] &= 6.1\bar{9}06\times 10^{6}~\mathrm{ng} \\[1.5ex] &= 6.19\times 10^{6}~\mathrm{ng} \end{align*}\]
Concept: moles; number of particles
What is the mass (in g) of 2.0 × 1024 mercury atoms?
Solution
Answer: 6.7 × 102 g
\[\begin{align*} m(\mathrm{Hg}) &= n(\mathrm{Hg}) ~ M(\mathrm{Hg}) \\[1.5ex] &= N(\mathrm{Hg}) ~ N_{\mathrm{A}}^{-1} ~ M(\mathrm{Hg}) \\[1.5ex] &= 2.0\times 10^{24}~\mathrm{Hg} \left ( \dfrac{\mathrm{mol}}{6.022\times 10^{23}~\mathrm{Hg}} \right ) \left ( \dfrac{200.59~\mathrm{g}}{\mathrm{mol}} \right )\\[1.5ex] &= 6.\bar{6}61\times 10^{2}~\mathrm{g} \\[1.5ex] &= 6.7\times 10^{2}~\mathrm{g} \end{align*}\]
Concept: moles; number of particles
What is the mass (in g) of 2.0 × 1024 mercury atoms?
Solution
Answer: 6.7 × 102 g
\[\begin{align*} m(\mathrm{Hg}) &= N(\mathrm{Hg}) ~ N_{\mathrm{A}}^{-1} ~ M(\mathrm{Hg}) \\[1.5ex] &= 2.0\times 10^{24}~\mathrm{Hg} \left ( \dfrac{\mathrm{mol}}{6.022\times 10^{23}~\mathrm{Hg}} \right ) \left ( \dfrac{200.59~\mathrm{g}}{\mathrm{mol}} \right )\\[1.5ex] &= 6.\bar{6}61\times 10^{2}~\mathrm{g} \\[1.5ex] &= 6.7\times 10^{2}~\mathrm{g} \end{align*}\]
Concept: moles; number of particles
The mineral spodumene has the formula LiAlSi2O6. How many lithium atoms are present in a 105 g sample?
Solution
Answer: 3.40 × 1023
\[\begin{align*} N(\mathrm{Li}) &= n(\mathrm{Li})~ N_{\mathrm{A}} \\[1.5ex] &= m(\mathrm{Li})~ M(\mathrm{Li})^{-1}~ N_{\mathrm{A}} \\[1.5ex] &= w(\mathrm{Li})~ m(\mathrm{sample})~ M(\mathrm{Li})^{-1}~ N_{\mathrm{A}} \\[1.5ex] &= \left ( \dfrac{m(\mathrm{Li})}{m(\mathrm{LiAlSi_2O_6})} \right ) m(\mathrm{sample})~ M(\mathrm{Li})^{-1}~ N_{\mathrm{A}} \\[1.5ex] &= \left ( \dfrac{A_{\mathrm{r}}^{\circ}(\mathrm{Li})} { A_{\mathrm{r}}^{\circ}(\mathrm{Li}) + A_{\mathrm{r}}^{\circ}(\mathrm{Al}) + \left [ 2~A_{\mathrm{r}}^{\circ}(\mathrm{Si}) \right ] + \left [ 6~A_{\mathrm{r}}^{\circ}(\mathrm{O}) \right ] } \right ) m(\mathrm{sample}) ~ M(\mathrm{Li})^{-1}~ N_{\mathrm{A}} \\[1.5ex] &= \left ( \dfrac{6.94} { 6.94 + 26.98 + \left [ 2\times 28.09 \right ] + \left [ 6\times 16.00 \right ] } \right ) \left ( \dfrac{105~\mathrm{g}}{} \right ) \left ( \dfrac{\mathrm{mol}}{\mathrm{6.94~\mathrm{g}}} \right ) \left ( \dfrac{6.022\times 10^{23}}{\mathrm{mol}} \right ) \\[1.5ex] &= \left ( \dfrac{6.94}{186.10}\right ) \left ( \dfrac{105~\mathrm{g}}{} \right ) \left ( \dfrac{\mathrm{mol}}{\mathrm{6.94~\mathrm{g}}} \right ) \left ( \dfrac{6.022\times 10^{23}}{\mathrm{mol}} \right ) \\[1.5ex] &= 3.3\bar{9}76\times 10^{23} \\[1.5ex] &= 3.40\times 10^{23} \end{align*}\]
Concept: number of particles
Find the mass (in g) of 500 atoms of iron.
Solution
Answer: 5 × 10–20 g
\[\begin{align*} m(\mathrm{Fe}) &= n(\mathrm{Fe}) ~ M(\mathrm{Fe}) \\[1.5ex] &= N(\mathrm{Fe}) ~ N_{\mathrm{A}}^{-1} ~ M(\mathrm{Fe}) \\[1.5ex] &= 500~\mathrm{Fe} \left ( \dfrac{\mathrm{mol}}{6.022\times 10^{23}~\mathrm{Fe}} \right ) \left ( \dfrac{55.85~\mathrm{g}}{\mathrm{mol}} \right ) \\[1.5ex] &= \bar{4}.63\times 10^{-20}~\mathrm{g} \\[1.5ex] &= 5\times 10^{-20}~\mathrm{g} \end{align*}\]
Concept: moles; number of particles
How much Fe (in mol and number of atoms) are in 125.0 g of Fe?
Solution
Answer: 2.238 mol; 1.347 × 1024 atoms
\[\begin{align*} n(\mathrm{Fe}) &= m(\mathrm{Fe}) ~ M(\mathrm{Fe})^{-1} \\[1.5ex] &= 125.0~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{55.85~\mathrm{g}} \right ) \\[1.5ex] &= 2.23\bar{8}13~\mathrm{mol} \\[1.5ex] &= 2.238~\mathrm{mol}\\[2.0ex] N(\mathrm{Fe}) &= n(\mathrm{Fe}) ~ N_{\mathrm{A}} \\[1.5ex] &= 2.23\bar{8}13~\mathrm{mol} \left ( \dfrac{6.022\times 10^{23}~\mathrm{Fe}} {\mathrm{mol}} \right ) \\[1.5ex] &= 1.34\bar{7}80\times 10^{24} \\[1.5ex] &= 1.347\times 10^{24} \end{align*}\]
Concept: number of particles
Freon-12 (CCl2F2) is used as a refrigerant in air conditioners and as a propellant in aerosol cans. What is the number of freon-12 molecules and what is the mass (in mg) of Cl in a 5.56 mg sample of freon-12?
Solution
Answer: 2.77 × 1019 molecules; 3.26 mg
\[\begin{align*} N(\mathrm{CCl_2F_2}) &= n(\mathrm{CCl_2F_2}) ~ N_{\mathrm{A}} \\[1.5ex] &= m(\mathrm{CCl_2F_2}) ~ M(\mathrm{CCl_2F_2})^{-1} ~ N_{\mathrm{A}} \\[1.5ex] &= 5.56~\mathrm{mg} \left ( \dfrac{\mathrm{g}}{10^3~\mathrm{mg}} \right ) \left ( \dfrac{\mathrm{mol}}{120.91~\mathrm{g}} \right ) \left ( \dfrac{6.022\times 10^{23}}{\mathrm{mol}} \right ) \\[1.5ex] &= 2.7\bar{6}91\times 10^{19}~\mathrm{CCl_2F_2} \\[1.5ex] &= 2.77\times 10^{19}~\mathrm{CCl_2F_2} \\[2.0ex] m(\mathrm{Cl}) &= n(\mathrm{Cl}) ~ M(\mathrm{Cl}) \\[1.5ex] &= N(\mathrm{CCl_2F_2}) ~ r(\mathrm{Cl,CCl_2F_2}) ~ N_{\mathrm{A}}^{-1} ~ M(\mathrm{Cl}) \\[1.5ex] &= 2.7\bar{6}91\times 10^{19}~\mathrm{CCl_2F_2} \left ( \dfrac{2~\mathrm{Cl}}{1~\mathrm{CCl_2F_2}} \right ) \left ( \dfrac{\mathrm{mol}}{6.022\times 10^{23}~\mathrm{Cl}} \right ) \left ( \dfrac{35.45~\mathrm{g}}{\mathrm{mol}} \right ) \left ( \dfrac{10^3~\mathrm{mg}}{\mathrm{g}} \right ) \\[2.0ex] &= 3.2\bar{6}01~\mathrm{mg} \\[1.5ex] &= 3.26~\mathrm{mg} \end{align*}\]
Concept: moles; number of particles
Prevacid (C16H14F3N3O2S) is used to treat gastroesophageal reflux disease (GERD). Determine each of the following:
- the molar mass (in g mol–1) of Prevacid
- the mass (in g) of fluorine in a 0.75 mol sample of Prevacid
- the number of C atoms in a 0.75 mol sample of Prevacid
- the mass (in g) of 4.25 × 1021 molecules of Prevacid
Solution
Answer:
- 369.36 g mol–1
- 43 g
- 7.2 × 1024
- 2.61 g
A. \[\begin{align*} M(\mathrm{PA}) &= \Biggl\{ \left [ 16~A_{\mathrm{r}}^{\circ}(\mathrm{C}) \right ] + \left [ 14~A_{\mathrm{r}}^{\circ}(\mathrm{H}) \right ] + \left [ 3~A_{\mathrm{r}}^{\circ}(\mathrm{F}) \right ] + \left [ 3~A_{\mathrm{r}}^{\circ}(\mathrm{N}) \right ] + \left [ 2~A_{\mathrm{r}}^{\circ}(\mathrm{O}) \right ] + A_{\mathrm{r}}^{\circ}(\mathrm{S}) \Biggl\} ~ M_{\mathrm{u}} \\[1.5ex] &= \Biggl\{ \left [ 16 \times 12.01 \right ] + \left [ 14 \times 1.01 \right ] + \left [ 3 \times 19.00 \right ] + \left [ 3 \times 14.01 \right ] + \left [ 2 \times 16.00 \right ] + 32.06 \Biggl\}~ \left ( \dfrac{\mathrm{g}}{\mathrm{mol}} \right ) \\[1.5ex] &= 369.39~\mathrm{g~mol^{-1}}\\[1.5ex] \end{align*}\]
B. \[\begin{align*} m(\mathrm{F}) &= n(\mathrm{F}) ~ M(\mathrm{F}) \\[1.5ex] &= n(\mathrm{PA})~ r(\mathrm{F,PA})~ M(\mathrm{F}) \\[1.5ex] &= 0.75~\mathrm{mol~PA} \left ( \dfrac{3~\mathrm{mol~F}}{\mathrm{mol~PA}} \right ) \left ( \dfrac{19.00~\mathrm{g}}{\mathrm{mol~F}} \right ) \\[1.5ex] &= 4\bar{2}.75~\mathrm{g} \\[1.5ex] &= 43~\mathrm{g} \end{align*}\]
C. \[\begin{align*} N(\mathrm{C}) &= n(\mathrm{C}) ~ N_{\mathrm{A}} \\[1.5ex] &= n(\mathrm{PA}) ~ r(\textrm{C,PA}) ~ N_{\mathrm{A}} \\[1.5ex] &= 0.75~\mathrm{mol~PA} \left ( \dfrac{16~\mathrm{mol~C}}{\mathrm{mol~PA}} \right ) %\\ % &\phantom{=}~ \left ( \dfrac{6.022\times 10^{23}~\mathrm{C}}{\mathrm{mol}} \right ) \\[1.5ex] &= 7.\bar{2}26\times 10^{24} = 7.2\times 10^{24}\\[3ex] \end{align*}\]
D. \[\begin{align*} m(\mathrm{PA}) &= n(\mathrm{PA}) ~ M(\mathrm{PA})\\[1.5ex] &= N(\mathrm{PA}) ~ N_{\mathrm{A}}^{-1}~ M(\mathrm{PA}) \\[1.5ex] &= 4.25\times 10^{21}~\mathrm{PA} \left ( \dfrac{\mathrm{mol}}{6.022\times 10^{23}~\mathrm{PA}} \right ) \left ( \dfrac{369.39~\mathrm{g}}{\mathrm{mol}} \right ) \\[1.5ex] &= 2.6\bar{0}69~\mathrm{g~PA} \\[1.5ex] &= 2.61~\mathrm{g~PA} \end{align*}\]
Concept: moles; number of particles
Find the percent composition by mass (to one decimal place) of each element in YBa2Cu3O7.
Solution
Answer: Y: 13.3 %; Ba: 41.2 %; Cu: 28.6 %; O: 16.8 %
Molar mass of YBa2Cu3O7:
\[\begin{align*} M(\mathrm{YBa_2Cu_3O_7}) &= \Biggl\{ A_{\mathrm{r}}^{\circ}(\mathrm{Y}) + \left [ 2~A_{\mathrm{r}}^{\circ}(\mathrm{Ba}) \right ] + \left [ 3~A_{\mathrm{r}}^{\circ}(\mathrm{Cu}) \right ] + \left [ 7~A_{\mathrm{r}}^{\circ}(\mathrm{O}) \right ] \Biggl\} ~ M_{\mathrm{u}} \\[1.5ex] &= \Biggl\{ 88.91 + \left [ 2 \times 137.33 \right ] + \left [ 3 \times 63.55 \right ] + \left [ 7 \times 16.00 \right ] \Biggl\} \left ( \dfrac{\mathrm{g}}{\mathrm{mol}} \right ) \\[1.5ex] &= 666.22~\mathrm{g~mol^{-1}} \end{align*}\]
Mass percent of each element in YBa2Cu3O7:
\[\begin{align*} w(\mathrm{Y})\% &= w(\mathrm{Y}) \times 100~\% \\[1.5ex] &= \left ( \dfrac{A_{\mathrm{r}}^{\circ}(\mathrm{Y})}{A_{\mathrm{r}}^{\circ}(\mathrm{YBa_2Cu_3O_7})} \right ) \times 100~\% \\[1.5ex] &= \left ( \dfrac{88.91}{666.22} \right ) \times 100~\% \\[1.5ex] &= 13.\bar{3}45~\% \\[1.5ex] &= 13.3~\%~\mathrm{Y}\\[3ex] w(\mathrm{Ba})\% &= w(\mathrm{Ba}) \times 100~\% \\[1.5ex] &= \left ( \dfrac{2~A_{\mathrm{r}}^{\circ}(\mathrm{Ba})}{A_{\mathrm{r}}^{\circ}(\mathrm{YBa_2Cu_3O_7})} \right ) \times 100~\% \\[1.5ex] &= \left ( \dfrac{2\times 137.33}{666.22} \right ) \times 100~\% \\[1.5ex] &= 41.\bar{2}26~\% \\[1.5ex] &= 41.2~\%~\mathrm{Ba} \\[3ex] w(\mathrm{Cu})\% &= w(\mathrm{Cu}) \times 100~\% \\[1.5ex] &= \left ( \dfrac{3~A_{\mathrm{r}}^{\circ}(\mathrm{Cu})}{A_{\mathrm{r}}^{\circ}(\mathrm{YBa_2Cu_3O_7})} \right ) \times 100~\% \\[1.5ex] &= \left ( \dfrac{3\times 63.55}{666.22} \right ) \times 100~\% \\[1.5ex] &= 28.\bar{6}16~\% \\[1.5ex] &= 28.6~\%~\mathrm{Cu} \\[3ex] w(\mathrm{O})\% &= w(\mathrm{O}) \times 100~\% \\[1.5ex] &= \left ( \dfrac{7~A_{\mathrm{r}}^{\circ}(\mathrm{O})}{A_{\mathrm{r}}^{\circ}(\mathrm{YBa_2Cu_3O_7})} \right ) \times 100~\% \\[1.5ex] &= \left ( \dfrac{7\times 16.00}{666.22} \right ) \times 100~\% \\[1.5ex] &= 16.\bar{8}11~\% \\[1.5ex] &= 16.8~\%~\mathrm{O} \end{align*}\]
Concept: percent composition by mass
Hemoglobin is a protein that transports oxygen in mammals. Hemoglobin is 0.347 % Fe (by mass). Each hemoglobin molecule contains 4 Fe atoms. What is the molar mass (in g mol–1 in standard notation) of hemoglobin?
Solution
Answer: 64 400 g mol–1
\[\begin{align*} w(\mathrm{Fe})\% &= w(\mathrm{Fe}) \times 100~\% \\[1.5ex] &= \left ( \dfrac{m(\mathrm{Fe})}{m(\mathrm{hemoglobin})} \right ) \times 100~\% \\[1.5ex] &= \dfrac{4~A_{\mathrm{r}}^{\circ}(\mathrm{Fe}) ~ M_{\mathrm{u}}} {\left ( \dfrac{w(\mathrm{Fe})\%}{100~\%}\right ) } \\[1.5ex] &= \dfrac{4~(55.85) \left ( \frac{\mathrm{g}}{\mathrm{mol}}\right ) } {\left ( \dfrac{0.347~\%}{100~\%}\right ) } \\[1.5ex] &= 64~\bar{3}80.4~\mathrm{g~mol^{-1}}\\[1.5ex] &= 64~400~\mathrm{g~mol^{-1}} \end{align*}\]
Concept: percent composition by mass
A compound that only contains carbon, hydrogen, and oxygen is 48.64 % C and 8.16 % H (by mass). What is the empirical formula of this substance?
Solution
Answer: C3H6O2
Assume an exact 100 g sample.
Find the moles of C, H, and O in an exact 100 g sample.
\[\begin{align*} n(\mathrm{C}) &= w(\mathrm{C}) ~ m(\mathrm{sample}) ~ M(\mathrm{C})^{-1} \\[1.5ex] &= \left ( \dfrac{w(\mathrm{C})\%}{100~\%} \right ) ~ m(\mathrm{sample}) ~ M(\mathrm{C})^{-1} \\[1.5ex] &= \left ( \dfrac{48.64\%~\mathrm{C}}{100~\%}\right ) \left ( \dfrac{100~\mathrm{g}}{} \right ) \left ( \dfrac{\mathrm{mol}}{\mathrm{12.01~g}} \right ) \\[1.5ex] &= 4.04\bar{9}95~\mathrm{mol~C}\\[3ex] n(\mathrm{H}) &= w(\mathrm{H}) ~ m(\mathrm{sample}) ~ M(\mathrm{H})^{-1} \\[1.5ex] &= \left ( \dfrac{w(\mathrm{H})\%}{100~\%} \right ) ~ m(\mathrm{sample}) ~ M(\mathrm{H})^{-1} \\[1.5ex] &= \left ( \dfrac{8.16\%~\mathrm{H}}{100~\%}\right ) \left ( \dfrac{100~\mathrm{g}}{} \right ) \left ( \dfrac{\mathrm{mol}}{\mathrm{1.01~g}} \right ) \\[1.5ex] &= 8.07\bar{9}20~\mathrm{mol~H}\\[3ex] n(\mathrm{O}) &= w(\mathrm{O}) ~ m(\mathrm{sample}) ~ M(\mathrm{O})^{-1} \\[1.5ex] &= \left ( \dfrac{w(\mathrm{O})\%}{100~\%} \right ) ~ m(\mathrm{sample}) ~ M(\mathrm{O})^{-1} \\[1.5ex] &= \left ( \dfrac{\left [ 100 - (48.64 + 8.16) \right ]~\%~\mathrm{O}}{100~\%}\right ) \left ( \dfrac{100~\mathrm{g}}{} \right ) \left ( \dfrac{\mathrm{mol}}{\mathrm{16.00~g}} \right ) \\[1.5ex] &= 2.70\bar{0}00~\mathrm{mol~O}\\[3ex] \end{align*}\]
Find the mole ratio of C, H, and O to a smallest whole number ratio by dividing by the smallest mole number and multiplying by an integer (here the integer is 2).
\[\begin{equation*} \begin{gathered} \mathrm{C}_{4.04\bar{9}95}\mathrm{H}_{8.07\bar{9}20}\mathrm{O}_{2.70\bar{0}00} \\[1.5ex] \mathrm{C}_{\frac{4.04\bar{9}9}{2.70\bar{0}00}}\mathrm{H}_{\frac{8.07\bar{9}2}{2.70\bar{0}00}}\mathrm{O}_{\frac{{2. 70\bar{0}00}}{2.70\bar{0}00}} \\[1.5ex] \mathrm{C}_{1.5}\mathrm{H}_{3}\mathrm{O}_{} \\[1.5ex] \mathrm{C}_{1.5\times 2}\mathrm{H}_{3\times 2}\mathrm{O}_{1\times 2} \\[4ex] \mathrm{C}_{3}\mathrm{H}_{6}\mathrm{O}_{2} \end{gathered} \end{equation*}\]
Concept: percent composition by mass; empirical formula
Consider four individual samples of phosphine (PH3), water, hydrogen sulfide, and hydrogen fluoride, each with a mass of 121 g. Rank the compounds from the least to the greatest number of hydrogen atoms contained in each sample.
Solution
Answer: HF < H2S < PH3 < H2O
PH3
\[\begin{align*} N(\mathrm{H}) &= n(\mathrm{PA}) ~ M(\mathrm{PA})\\[1.5ex] &= m(\mathrm{PH_3}) ~ M(\mathrm{PH_3})^{-1} ~ r(\mathrm{H,PH_3}) ~ N_{\mathrm{A}} \\[1.5ex] &= 121~\mathrm{g~PH_3} \left ( \dfrac{\mathrm{mol~PH_3}}{34.00~\mathrm{g}} \right ) \left ( \dfrac{3~\mathrm{mol~H}}{\mathrm{mol~PH_3}} \right ) \left ( \dfrac{6.022\times 10^{23}~\mathrm{H}}{\mathrm{mol~H}} \right ) \\[1.5ex] &= 6.4\bar{2}93\times 10^{24}~\mathrm{H} \\[1.5ex] &= 6.43\times 10^{24}~\mathrm{H} \end{align*}\]
H2O
\[\begin{align*} N(\mathrm{H}) &= n(\mathrm{PA}) ~ M(\mathrm{PA})\\[1.5ex] &= m(\mathrm{H_2O}) ~ M(\mathrm{H_2O})^{-1} ~ r(\mathrm{H,H_2O}) ~ N_{\mathrm{A}} \\[1.5ex] &= 121~\mathrm{g~H_2O} \left ( \dfrac{\mathrm{mol~H_2O}}{18.02~\mathrm{g}} \right ) \left ( \dfrac{2~\mathrm{mol~H}}{\mathrm{mol~H_2O}} \right ) \left ( \dfrac{6.022\times 10^{23}~\mathrm{H}}{\mathrm{mol~H}} \right ) \\[1.5ex] &= 8.0\bar{8}72\times 10^{24}~\mathrm{H} \\[1.5ex] &= 8.09\times 10^{24}~\mathrm{H} \end{align*}\]
H2S
\[\begin{align*} N(\mathrm{H}) &= n(\mathrm{PA}) ~ M(\mathrm{PA})\\[1.5ex] &= m(\mathrm{H_2S}) ~ M(\mathrm{H_2S})^{-1} ~ r(\mathrm{H,H_2S}) ~ N_{\mathrm{A}} \\[1.5ex] &= 121~\mathrm{g~H_2S} \left ( \dfrac{\mathrm{mol~H_2S}}{34.08~\mathrm{g}} \right ) \left ( \dfrac{2~\mathrm{mol~H}}{\mathrm{mol~H_2S}} \right ) \left ( \dfrac{6.022\times 10^{23}~\mathrm{H}}{\mathrm{mol~H}} \right ) \\[1.5ex] &= 4.2\bar{7}61\times 10^{24}~\mathrm{H} \\[1.5ex] &= 4.28\times 10^{24}~\mathrm{H} \end{align*}\]
HF
\[\begin{align*} N(\mathrm{H}) &= n(\mathrm{PA}) ~ M(\mathrm{PA})\\[1.5ex] &= m(\mathrm{HF}) ~ M(\mathrm{HF})^{-1} ~ r(\mathrm{H,HF}) ~ N_{\mathrm{A}} \\[1.5ex] &= 121~\mathrm{g~HF} \left ( \dfrac{\mathrm{mol~HF}}{20.01~\mathrm{g}} \right ) \left ( \dfrac{1~\mathrm{mol~H}}{\mathrm{mol~HF}} \right ) \left ( \dfrac{6.022\times 10^{23}~\mathrm{H}}{\mathrm{mol~H}} \right ) \\[1.5ex] &= 3.6\bar{4}14\times 10^{24}~\mathrm{H} \\[1.5ex] &= 3.64\times 10^{24}~\mathrm{H} \end{align*}\]
Concept: moles; number of particles
Balancing Equations, Solubility
Write a balanced equation for the following reaction by placing appropriate stoichiometric coefficients.
\[\begin{align*} \rule[-1.0pt]{2em}{0.5pt} \mathrm{CH_3OH} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{O_2} ~\longrightarrow~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{CO_2} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{H_2O} \end{align*}\]
Solution
Answer:
\[\begin{align*} \mathrm{2~CH_3OH} + \mathrm{3~O_2} \longrightarrow \mathrm{2~CO_2} + \mathrm{4~H_2O} \end{align*}\]
Concept: balancing equations
Write a balanced equation for the following reaction by placing appropriate stoichiometric coefficients.
\[\begin{align*} \rule[-1.0pt]{2em}{0.5pt} \mathrm{CH_3NHNH_2} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{O_2} ~\longrightarrow~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{CO_2} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{H_2O} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{H_2} \end{align*}\]
Solution
Answer:
\[\begin{align*} \mathrm{2~CH_3OH} + \mathrm{5~O_2} \longrightarrow \mathrm{2~CO_2} + \mathrm{6~H_2O} + \mathrm{2~N_2} \end{align*}\]
Concept: balancing equations
Write a balanced equation for the following reaction by placing appropriate stoichiometric coefficients.
\[\begin{align*} \rule[-1.0pt]{2em}{0.5pt} \mathrm{Se} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{BrF_5} ~\longrightarrow~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{SeF_6} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{BrF_3} \end{align*}\]
Solution
Answer:
\[\begin{align*} \mathrm{1~Se} + \mathrm{3~BrF_5} \longrightarrow \mathrm{2~SeF_6} + \mathrm{4~BrF_3} \end{align*}\]
Concept: balancing equations
Determine the proper value for m and n needed to balance each equation.
\[\begin{align*} \mathrm{B}_m\mathrm{H}_n + \mathrm{3~O_2} \longrightarrow \mathrm{B_2O_3} + \mathrm{3~H_2O} \end{align*}\]
Solution
Answer: m = 2; n = 6
Concept: balancing equations
\[\begin{align*} \mathrm{B_2H_6} + \mathrm{3~O_2} \longrightarrow \mathrm{B_2O_3} + \mathrm{3~H_2O} \end{align*}\]
Determine the proper value for m and n needed to balance each equation.
\[\begin{align*} \mathrm{H}_m\mathrm{IO}_n ~\longrightarrow~ \mathrm{H^+} + \mathrm{IO_4^-} + \mathrm{2~H_2O} \end{align*}\]
Solution
Answer: m = 5; n = 6
Concept: balancing equations
\[\begin{align*} \mathrm{H_5IO_6} \longrightarrow \mathrm{H^+} + \mathrm{IO_4^-} + \mathrm{2~H_2O} \end{align*}\]
Write a balanced equation for the following reaction by placing appropriate stoichiometric coefficients.
\[\begin{align*} \rule[-1.0pt]{2em}{0.5pt} \mathrm{CH_4} + \rule[-1.0pt]{2em}{0.5pt} \mathrm{H_2O} ~\longrightarrow~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{CO} + \rule[-1.0pt]{2em}{0.5pt} \mathrm{H_2} \end{align*}\]
Solution
Answer:
\[\begin{align*} \mathrm{1~CH_4} + \mathrm{1~H_2O} \longrightarrow \mathrm{1~CO} + \mathrm{3~H_2} \end{align*}\]
Concept: balancing equations
Write a balanced equation for the following reaction by placing appropriate stoichiometric coefficients.
\[\begin{align*} \rule[-1.0pt]{2em}{0.5pt} \mathrm{Ag_2O(s)} ~\longrightarrow~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{Ag(s)} + \rule[-1.0pt]{2em}{0.5pt} \mathrm{O_2} \end{align*}\]
Solution
Answer:
\[\begin{align*} \mathrm{2~Ag_2O(s)} \longrightarrow \mathrm{4~Ag(s)} + \mathrm{1~O_2} \end{align*}\]
Concept: balancing equations
Chemical equations must be balanced because the resulting coefficients allow us to predict (select all that apply).
- the amount of product that can form from a given amount of reactant.
- whether the reaction requires a catalyst or not
- how much of one reactant is required to react with a given amount of another
- how much reactants are required to form a given amount of products
- whether the given reaction is possible or not
Solution
Answer: A, C, D
Concept: balancing equations
Chemical equations must be balanced because the resulting coefficients allow us to predict (select all that apply).
aqueous silver sulfate + aqueous barium iodide → solid barium sulfate + solid silver iodide- Ag2SO4 + BaI2 → BaSO4 + AgI
- Ag2SO4(l) + BaI2(l) → BaSO4(s) + 2 AgI(s)
- Ag2SO4(aq) + BaI2(aq) → BaSO4(s) + 2 AgI(s)
- AgSO4(aq) + BaI(aq) → BaSO4(s) + 2 AgI(s)
- AgSO4(l) + 2 BaI(l) → Ba2SO4(s) + 2 AgI2(s)
Solution
Answer: C
Concept: balancing equations; nomenclature
Classify each of the following (all that apply) as a
I. strong electrolyte
II. weak electrolyte
III. nonelectrolyte
IV. strong acid
V. strong base
VI. weak acid
VII. weak base
VIII. ionic compound
IX. organic compound
- HBr
- ammonium carbonate
- NaClO4
- ethanol
- acetic acid
- NH3
Solution
Answer:
- I, IV
- I, VIII
- I, VIII
- III, IX
- II, VI
- II, VI
Concept: electrolytes
Which of the following compounds are insoluble in water? Select all that apply.
- CoCO3
- Cu3(PO4)2
- AgNO3
- Na2S
- AgI
Solution
Answer: A, B, E
Concept: solubility rules
Which of the following combinations will form a precipitate? Select all that apply.
- SrCl2(aq) + Na2S(aq)
- KCl(aq) + CaS(aq)
- Hg(NO3)2(aq) + Na3PO4(aq)
- Ba (NO3)2(aq) + KOH (aq)
- NaOH(aq) + FeCl3(aq)
Solution
Answer: A, C, E
Concept: solubility rules; precipitation
- SrCl2(aq) + Na2S(aq) → SrS(s) + 2 NaCl(aq)
- 2 KCl(aq) + CaS(aq) → K2S(aq) + CaCl2(aq)
- Hg(NO3)2(aq) + 2 Na3PO4(aq) → Hg3(PO4)2(s) + 6 NaNO3(aq)
- Ba(NO3)2(aq) + KOH(aq) → Ba(OH)2(aq) + 2 KNO3
Lead(II) nitrate reacts with sodium chloride. Choose the net ionic equation for the reaction.
- Pb2+(aq) + Cl–(aq) → PbCl(s)
- Pb2+(aq) + NO3–(aq) + Na+(aq) + Cl–(aq) → PbCl(s) + NaNO3(aq)
- PbNO3(aq) + NaCl(aq) → PbCl(s) + NaNO3(aq)
- Pb2+(aq) + 2 Cl–(aq) → PbCl2(s)
Solution
Answer: D
Concept: solubility rules; precipitation
Molecular equation:
Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaNO3(aq)
Complete ionic equation:
Pb2+(aq) + 2 NO3–(aq) + 2 Na+(aq) + 2 Cl–(aq) → PbCl2(s) + 2 Na+(aq) + 2NO3–(aq)
Net ionic equation:
Pb2+(aq) + 2 Cl–(aq) → PbCl2(s)
Aqueous solutions of sodium sulfate and barium chloride react. What is the sum of the coefficients from the balanced net ionic equation?
Solution
Answer: 3
Concept: solubility rules; precipitation; net ionic equation; balancing equations
Molecular equation:
Na2SO4(aq) + BaCl2(aq) → 2 NaCl(aq) + BaSO4(s)
Complete ionic equation:
2 Na2+(aq) + SO42–(aq) + Ba2+(aq) + 2 Cl–(aq) → 2 Na + (aq) + 2 Cl–(aq) + BaSO4(s)
Net ionic equation:
Ba2+(aq) + SO42–(aq) → BaSO4(s)
Answer the questions for the following reaction.
2 HI(aq) + Ca(OH)2(aq) → 2 H2O(l) + CaI2(aq)- Is the acid strong or weak?
- Is the base strong or weak?
- What is the net ionic equation for the reaction?
Solution
Answer:
- strong
- strong
- H+(aq) + OH–(aq) → H2O(l)
Concept: solubility rules; precipitation; net ionic equation; balancing equations
Molecular equation:
2 HI(aq) + Ca(OH)2(aq) → 2 H2O(l) + CaI2(aq)
Complete ionic equation:
2 H+(aq) + 2 I–(aq) + Ca2+(aq) + 2 OH–(aq) → 2 H2O(l) + Ca2+(aq) + 2 I–(aq)
Net ionic equation:
H+(aq) + OH–(aq) → H2O(l)
A reaction between hydrobromic acid and potassium hydroxide occurs. What is the sum of the coefficients from the balanced net ionic equation?
- Is the acid strong or weak?
- Is the base strong or weak?
- What is the net ionic equation for the reaction?
Solution
Answer: 3
Concept: acid-base reaction; net ionic equation; balancing equations
Molecular equation:
HBr(aq) + KOH(aq) → KBr(aq) + H2O(l)
Complete ionic equation:
H+(aq) + Br–(aq) + K+(aq) + OH–(aq) → K+(aq) + Br–(aq) + H2O(l)
Net ionic equation:
H+(aq) + OH–(aq) → H2O(l)
Which is the spectator ion in the reaction between potassium carbonate and calcium iodide? Select all that apply.
- K+(aq)
- CO32–(aq)
- Ca2+(aq)
- I–(aq)
Solution
Answer: A, D
Concept: solubility rules; precipitation; net ionic equation; balancing equations
Molecular equation:
K2CO3(aq) + CaI2(aq) → 2 KI(aq) + CaCO3(s)
Complete ionic equation:
2 K+(aq) + CO32–(aq) + Ca2+(aq) + 2 I–(aq) → 2 K+(aq) + 2 I–(aq) + CaCO3(s)
Net ionic equation:
Ca2+(aq) + CO32–(aq) → CaCO3(s)
What is the sum of the coefficients of the net ionic equation for aqueous sodium hydroxide neutralized by aqueous acetic acid?
Solution
Answer: 4
Concept: acid-base reaction; net ionic equation; balancing equations
Molecular equation:
CH3COOH(aq) + NaOH(aq) → NaCH3COO(aq) + H2O(l)
Complete ionic equation:
CH3COOH(aq) + Na+(aq) + OH–(aq) → Na+(aq) + CH3COO–(aq) + H2O(l)
Net ionic equation:
CH3COOH(aq) + OH–(aq) → CH3COO–(aq) + H2O(l)
When the following solutions are mixed together, what precipitate (if any) will form?
- FeSO4(aq) + KCl(aq)
- Al(NO3)3(aq) + Ba(OH)2(aq)
- CaCl2(aq) + Na2SO4(aq)
- K2S(aq) + Ni(NO3)2(aq)
- Hg2(NO3)2(aq) + CuSO4(aq)
- Ni(NO3)2(aq) + CaCl2(aq)
- K2CO3(aq) + MgI2(aq)
- Na2CrO4(aq) + AlBr3(aq)
Solution
Answer:
- FeSO4(aq) + KCl(aq) → FeCl2(aq) + K2SO4(aq) – none
- Al(NO3)3(aq) + Ba(OH)2(aq) → Al(OH)3(s) + Ba(OH)2(aq)
- CaCl2(aq) + Na2SO4(aq) → CaSO4(s) + NaCl(aq)
- K2S(aq) + Ni(NO3)2(aq) → KNO3(aq) + NiS(s)
- Hg2(NO3)2(aq) + CuSO4(aq) → Hg2SO4(s) + Cu(NO3)2(aq)
- Ni(NO3)2(aq) + CaCl2(aq) → NiCl2(aq) + Ca(NO3)2(aq) – none
- K2CO3(aq) + MgI2(aq) → KI(aq) + MgCO3(s)
- Na2CrO4(aq) + AlBr3(aq) → NaBr(aq) + Al2(CrO4)3(s)
Concept: solubility rules; double-displacement reaction
Which of the following substances are soluble in water? Select all that apply.
- aluminum nitrate
- magnesium chloride
- rubidium sulfate
- nickel(II) hydroxide
- lead(II) sulfide
- barium hydroxide
- iron(III) phosphate
Solution
Answer: A, B, C, F
Concept: solubility rules
Write the net ionic equations for the following reactions:
- ammonium sulfate and barium nitrate
- lead(II) nitrate and sodium chloride
- sodium phosphate and potassium nitrate
- sodium bromide and rubidium chloride
- copper(II) chloride and sodium hydroxide
Solution
Answer:
- Ba2+(aq) + SO42–(aq) → BaSO4(s)
- Pb2+(aq) + 2 Cl–(aq) → PbCl2(s)
- none (all ions are spectators)
- none (all ions are spectators)
- Cu2+(aq) + 2 NaOH(aq) → Cu(OH)2(s)
Concept: net ionic equation; balancing equations; solubility rules
Write the balanced molecular equation, complete ionic equation, and net ionic equation for the following acid-base reactions.
- HClO4(aq) + Mg(OH)2(s)
- HCN(aq) + NaOH(aq)
- HCl(aq) + NaOH(aq)
Solution
Answer:
A
Molecular equation
\[\begin{align*} 2~\mathrm{HClO_4(aq)} + \mathrm{Mg(OH)_2(s)} &\longrightarrow 2~\mathrm{H_2O(l)} + \mathrm{Mg(ClO_4)_2(aq)} \end{align*}\]
Complete ionic equation
\[\begin{align*} \mathrm{2~H^+(aq)} + \mathrm{2~ClO_4^-(aq)} + \mathrm{Mg(OH)_2(s)} &\longrightarrow \mathrm{2~H_2O(l)} + \mathrm{Mg^{2+}(aq)} + 2~\mathrm{ClO_4^-(aq)} \end{align*}\]
Net ionic equation
\[\begin{align*} \mathrm{2~H^{+}(aq)} + \mathrm{Mg(OH)_2(s)} &\longrightarrow \mathrm{2~H_2O(l)} + \mathrm{Mg^{2+}(aq)} \end{align*}\]
B
Molecular equation
\[\begin{align*} \mathrm{HCN(aq)} + \mathrm{NaOH(aq)} &\longrightarrow \mathrm{H_2O(l)} + \mathrm{NaCN(aq)} \end{align*}\]
Complete ionic equation
\[\begin{align*} \mathrm{H^+(aq)} + \mathrm{CN^-(aq)} + \mathrm{Na^+(aq)} + \mathrm{OH^-(aq)} &\longrightarrow \mathrm{H_2O(l)} + \mathrm{Na^{+}(aq)} + \mathrm{CN^-(aq)} \end{align*}\]
Net ionic equation
\[\begin{align*} \mathrm{H^{+}(aq)} + \mathrm{OH^-(aq)} &\longrightarrow \mathrm{H_2O(l)}$ \end{align*}\]
C
Molecular equation
\[\begin{align*} \mathrm{HCl(aq)} + \mathrm{NaOH(aq)} &\longrightarrow \mathrm{H_2O(l)} + \mathrm{NaCl(aq)} \end{align*}\]
Complete ionic equation
\[\begin{align*} \mathrm{H^+(aq)} + \mathrm{Cl^-(aq)} + \mathrm{Na^+(aq)} + \mathrm{OH^-(aq)} &\longrightarrow \mathrm{H_2O(l)} + \mathrm{Na^{+}(aq)} + \mathrm{Cl^-(aq)} \end{align*}\]
Net ionic equation
\[\begin{align*} \mathrm{H^{+}(aq)} + \mathrm{OH^-(aq)} &\longrightarrow \mathrm{H_2O(l)} \end{align*}\]
Concept: acid-base reaction; net ionic equation
Write a balanced chemical equation between an acid and a base that would have the following salt appear as a product.
- potassium perchlorate
- cesium nitrate
- calcium iodide
Solution
Answer:
A
\[\begin{align*} \mathrm{HClO_4(aq)} + \mathrm{KOH(aq)} &\longrightarrow \mathrm{H_2O(l)} + \mathrm{KClO_4(aq)} \end{align*}\]
B
\[\begin{align*} \mathrm{HNO_3(aq)} + \mathrm{CsOH(aq)} &\longrightarrow \mathrm{H_2O(l)} + \mathrm{CsNO_3(aq)} \end{align*}\]
C
\[\begin{align*} \mathrm{2~HI(aq)} + \mathrm{Ca(OH)_2(aq)} &\longrightarrow \mathrm{2~H_2O(l)} + \mathrm{CaI_2(aq)} \end{align*}\]
Concept: acid-base reaction; salts
Oxidation-reduction
Assign oxidation states for all atoms in each of the following compounds.
- KMnO4
- NiO2
- Na4Fe(OH)6
- (NH4)2HPO4
- P4O6
Solution
Answer:
A
\[\begin{align*} &\underset{\begin{subarray}{c} +1 \\ +1 \end{subarray}}{\phantom{_1}\mathrm{K}\phantom{_1}} \underset{\begin{subarray}{c} +7 \\ +7 \end{subarray}}{\phantom{_1}\mathrm{Mn}\phantom{_1}} \underset{\begin{subarray}{c} -2 \\ -8 \end{subarray}}{\phantom{_1}\mathrm{O_4}\phantom{_1}} \end{align*}\]
B
\[\begin{align*} &\underset{\begin{subarray}{c} +4 \\ +4 \end{subarray}}{\phantom{_1}\mathrm{Ni}\phantom{_1}} \underset{\begin{subarray}{c} -2 \\ -4 \end{subarray}}{\phantom{_1}\mathrm{O_2}\phantom{_1}} \end{align*}\]
C
\[\begin{align*} &\underset{\begin{subarray}{c} +1 \\ +4 \end{subarray}}{\phantom{_1}\mathrm{Na_4}\phantom{_1}} \underset{\begin{subarray}{c} +2 \\ +2 \end{subarray}}{\phantom{_1}\mathrm{Fe}\phantom{_1}} \underset{\begin{subarray}{c} -2 \\ -12 \end{subarray}}{(\phantom{_1}\mathrm{O}\phantom{_1}} \underset{\begin{subarray}{c} +1 \\ +6 \end{subarray}}{\phantom{_1}\mathrm{H}\phantom{_1})_6} \end{align*}\]
D
\[\begin{align*} &\underset{\begin{subarray}{c} -3 \\ -3 \end{subarray}}{( \phantom{_1}\mathrm{N}\phantom{_1}} \underset{\begin{subarray}{c} +1 \\ +4 \end{subarray}}{\phantom{_1}\mathrm{H_4}\phantom{_1}})_2 \underset{\begin{subarray}{c} +1 \\ +1 \end{subarray}}{\phantom{_1}\mathrm{H}\phantom{_1}} \underset{\begin{subarray}{c} +5 \\ +5 \end{subarray}}{\phantom{_1}\mathrm{P}\phantom{_1}} \underset{\begin{subarray}{c} -2 \\ -8 \end{subarray}}{\phantom{_1}\mathrm{O_4}\phantom{_1}} \end{align*}\]
E
\[\begin{align*} &\underset{\begin{subarray}{c} +3 \\ +12 \end{subarray}}{\phantom{_1}\mathrm{P_4}\phantom{_1}} \underset{\begin{subarray}{c} -2 \\ -12 \end{subarray}}{\phantom{_1}\mathrm{O_6}\phantom{_1}} \end{align*}\]
Concept: oxidation state
Assign oxidation states for all atoms in each of the following compounds.
- Fe3O4
- XeOF4
- SF4
- CO
- C6H12O6
Solution
Answer:
A
\[\begin{align*} &\underset{\begin{subarray}{c} +2/+3 \\ +8 \end{subarray}}{\phantom{_1}\mathrm{Fe_3}\phantom{_1}} \underset{\begin{subarray}{c} -2 \\ -8 \end{subarray}}{\phantom{_1}\mathrm{O_4}\phantom{_1}} \end{align*}\]
B
\[\begin{align*} &\underset{\begin{subarray}{c} +6 \\ +6 \end{subarray}}{\phantom{_1}\mathrm{Xe}\phantom{_1}} \underset{\begin{subarray}{c} -2 \\ -2 \end{subarray}}{\phantom{_1}\mathrm{O}\phantom{_1}} \underset{\begin{subarray}{c} -1 \\ -4 \end{subarray}}{\phantom{_1}\mathrm{F_4}\phantom{_1}} \end{align*}\]
C
\[\begin{align*} &\underset{\begin{subarray}{c} +4 \\ +4 \end{subarray}}{\phantom{_1}\mathrm{S}\phantom{_1}} \underset{\begin{subarray}{c} -1 \\ -4 \end{subarray}}{\phantom{_1}\mathrm{F_4}\phantom{_1}} \end{align*}\]
D
\[\begin{align*} &\underset{\begin{subarray}{c} +2 \\ +2 \end{subarray}}{\phantom{_1}\mathrm{C}\phantom{_1}} \underset{\begin{subarray}{c} -2 \\ -2 \end{subarray}}{\phantom{_1}\mathrm{O}\phantom{_1}} \end{align*}\]
E
\[\begin{align*} &\underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{\phantom{_1}\mathrm{C_6}\phantom{_1}} \underset{\begin{subarray}{c} +1 \\ +12 \end{subarray}}{\phantom{_1}\mathrm{H_{12}}\phantom{_1}} \underset{\begin{subarray}{c} -2 \\ -12 \end{subarray}}{\phantom{_1}\mathrm{O_6}\phantom{_1}} \end{align*}\]
Concept: oxidation state
Specify which reactions are redox reactions and identify the oxidizing agent, reducing agent, the substance being oxidized, and the substance being reduced.
- Cu(s) + 2 Ag+(aq) → 2 Ag(s) + Cu2+(aq)
- HCl(g) + NH3(g) → NH4Cl(s)
- SiCl4(l) + 2 H2O(l) → 4 HCl(aq) + SiO2(s)
- SiCl4(l) + 2 Mg(s) → 2 MgCl2(s) + Si(s)
- Al(OH)4–(aq) → AlO2–(aq) + 2 H2O(l)
Solution
Answer:
A
\[\begin{align*} \underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{\phantom{_1\!}\mathrm{Cu}\phantom{_1\!}}\mathrm{(s)} ~+~ \underset{\begin{subarray}{c} +1 \\ +1 \end{subarray}}{\phantom{_1\!}\mathrm{2~Ag^+}\phantom{_1\!}}\mathrm{(aq)} \longrightarrow \underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{\phantom{_1\!}\mathrm{2~Ag}\phantom{_1\!}}\mathrm{(s)} ~+~ \underset{\begin{subarray}{c} +2 \\ +2 \end{subarray}}{\phantom{_1\!}\mathrm{Cu^{2+}}\phantom{_1\!}}\mathrm{(aq)} \end{align*}\]
Oxidizing agent: Ag+(aq); Reducing agent: Cu(s)
Substance oxidized: Cu(s); Substance reduced: Ag+(aq)B
\[\begin{align*} \underset{\begin{subarray}{c} +1 \\ +1 \end{subarray}}{\phantom{_1\!}\mathrm{H}\phantom{_1\!}} \underset{\begin{subarray}{c} -1 \\ -1 \end{subarray}}{\phantom{_1\!}\mathrm{Cl}\phantom{_1\!}}\mathrm{(g)} ~+~ \underset{\begin{subarray}{c} -3 \\ -3 \end{subarray}}{\phantom{_1\!}\mathrm{N}\phantom{_1\!}} \underset{\begin{subarray}{c} +1 \\ +3 \end{subarray}}{\phantom{_1\!}\mathrm{H_3}\phantom{_1\!}}\mathrm{(g)} \longrightarrow \underset{\begin{subarray}{c} -3 \\ -3 \end{subarray}}{\phantom{_1\!}\mathrm{N}\phantom{_1\!}} \underset{\begin{subarray}{c} +1 \\ +4 \end{subarray}}{\phantom{_1\!}\mathrm{H_4}\phantom{_1\!}} \underset{\begin{subarray}{c} -1 \\ -1 \end{subarray}}{\phantom{_1\!}\mathrm{Cl}\phantom{_1\!}}\mathrm{(s)} \end{align*}\]
Not a redox reaction.
C
\[\begin{align*} \underset{\begin{subarray}{c} +4 \\ +4 \end{subarray}}{\phantom{_1\!}\mathrm{Si}\phantom{_1\!}} \underset{\begin{subarray}{c} -1 \\ -4 \end{subarray}}{\phantom{_1\!}\mathrm{Cl_4}\phantom{_1\!}}\mathrm{(l)} ~+~ \underset{\begin{subarray}{c} +1 \\ +2 \end{subarray}}{2~\phantom{_1\!}\mathrm{H_2}\phantom{_1\!}} \underset{\begin{subarray}{c} -2 \\ -2 \end{subarray}}{\phantom{_1\!}\mathrm{O}\phantom{_1\!}}\mathrm{(l)} \longrightarrow \underset{\begin{subarray}{c} +1 \\ +1 \end{subarray}}{4~\phantom{_1\!}\mathrm{H}\phantom{_1\!}} \underset{\begin{subarray}{c} -1 \\ -1 \end{subarray}}{\phantom{_1\!}\mathrm{Cl}\phantom{_1\!}}\mathrm{(aq)} ~+~ \underset{\begin{subarray}{c} +4 \\ +4 \end{subarray}}{\phantom{_1\!}\mathrm{Si}\phantom{_1\!}} \underset{\begin{subarray}{c} -2 \\ -4 \end{subarray}}{\phantom{_1\!}\mathrm{O_2}\phantom{_1\!}}\mathrm{(s)} \end{align*}\]
Not a redox reaction.
D
\[\begin{align*} \underset{\begin{subarray}{c} +4 \\ +4 \end{subarray}}{\phantom{_1\!}\mathrm{Si}\phantom{_1\!}} \underset{\begin{subarray}{c} -1 \\ -4 \end{subarray}}{\phantom{_1\!}\mathrm{Cl_4}\phantom{_1\!}}\mathrm{(l)} ~+~ \underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{2~\phantom{_1\!}\mathrm{Mg}\phantom{_1\!}}\mathrm{(s)} \longrightarrow \underset{\begin{subarray}{c} +2 \\ +2 \end{subarray}}{2~\phantom{_1\!}\mathrm{Mg}\phantom{_1\!}} \underset{\begin{subarray}{c} -1 \\ -2 \end{subarray}}{\phantom{_1\!}\mathrm{Cl_2}\phantom{_1\!}}\mathrm{(s)} ~+~ \underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{\phantom{_1\!}\mathrm{Si}\phantom{_1\!}}\mathrm{(s)} \end{align*}\]
Oxidizing agent: SiCl4(l); Reducing agent: Mg(s)
Substance oxidized: Mg(s); Substance reduced: SiCl4(l)E
\[\begin{align*} \underset{\begin{subarray}{c} +3 \\ +3 \end{subarray}}{\phantom{_1\!}\mathrm{Al}\phantom{_1\!}} \underset{\begin{subarray}{c} -2 \\ -8 \end{subarray}}{(\phantom{_1\!}\mathrm{O}\phantom{_1\!}} \underset{\begin{subarray}{c} +1 \\ +4 \end{subarray}}{\phantom{_1\!}\mathrm{H}\phantom{_1\!}})_4^- \mathrm{(aq)} \longrightarrow \underset{\begin{subarray}{c} +3 \\ +3 \end{subarray}}{2~\phantom{_1\!}\mathrm{Al}\phantom{_1\!}} \underset{\begin{subarray}{c} -2 \\ -4 \end{subarray}}{\phantom{_1\!}\mathrm{O_2}\phantom{_1\!}}^-\mathrm{(aq)} ~+~ \underset{\begin{subarray}{c} +1\\ +2 \end{subarray}}{2~\phantom{_1\!}\mathrm{H_2}\phantom{_1\!}} \underset{\begin{subarray}{c} -2\\ -2 \end{subarray}}{\phantom{_1\!}\mathrm{O}\phantom{_1\!}}\mathrm{(l)} \end{align*}\]
Not a redox reaction.
Concept: oxidation state
Determine the half-reactions, the amount (in mol) of electrons transferred, and the overall balanced reaction (with phase labels) for the following redox reaction.
\[\begin{align*} \mathrm{Fe(s)} + \mathrm{Br_2(l)} \longrightarrow \mathrm{FeBr_3(s)} \end{align*}\]
Solution
Answer:
\[\begin{align*} \underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{\phantom{_1\!}\mathrm{Fe}\phantom{_1\!}}\mathrm{(s)} ~+~ \underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{\phantom{_1\!}\mathrm{Br_2}\phantom{_1\!}}\mathrm{(l)} \longrightarrow \underset{\begin{subarray}{c} +3 \\ +3 \end{subarray}}{\phantom{_1\!}\mathrm{Fe}\phantom{_1\!}} \underset{\begin{subarray}{c} -1 \\ -3 \end{subarray}}{\phantom{_1\!}\mathrm{Br_3}\phantom{_1\!}}\mathrm{(s)} \end{align*}\]
Fe undergoes oxidation and Br2 undergoes reduction.
\[\begin{alignat*}{3} \textrm{Ox. half-reaction} &\quad & \mathrm{Fe} &\; &\longrightarrow &\; \mathrm{Fe^{3+}} + 3~e^- \\ \textrm{Red. half-reaction} &\quad & \mathrm{Br_2} + 2~e^- &\; &\longrightarrow &\; 3~\mathrm{Br^-}\\[1.0ex] \textit{balance e}^- &\quad & 2~( \mathrm{Fe} &\; &\longrightarrow &\; \mathrm{Fe^{3+}} + 3~e^- ) \\ &\quad & 3~( \mathrm{Br_2} + 2~e^- &\; &\longrightarrow &\; 3~\mathrm{Br^-})\\[1.0ex] \textit{6 mol e}^- &\quad & 2~\mathrm{Fe} &\; &\longrightarrow &\; 2~\mathrm{Fe^{3+}} + 6~e^- \\ &\quad & 3~\mathrm{Br_2} + 6~e^- &\; &\longrightarrow &\; 6~\mathrm{Br^-} \\[1.0ex] \textit{combine} &\quad & 2~\mathrm{Fe} + 3~\mathrm{Br_2} &\; &\longrightarrow &\; 2~\mathrm{Fe^{3+}} + 6~\mathrm{Br^-} \\[1.0ex] \textrm{Balanced reaction} &\quad & 2~\mathrm{Fe(s)} + 3~\mathrm{Br_2(l)} &\; &\longrightarrow &\; 2~\mathrm{FeBr_3(s)} \end{alignat*}\]
Concept: oxidation state; oxidation-reduction reaction; balancing equations
Determine the half-reactions, the amount (in mol) of electrons transferred, and the overall balanced reaction (with phase labels) for the following redox reaction.
\[\begin{align*} \mathrm{Mn(s)} + \mathrm{F_2(g)} \longrightarrow \mathrm{MnF_4(s)} \end{align*}\]
Solution
Answer:
\[\begin{align*} \underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{\phantom{_1\!}\mathrm{Mn}\phantom{_1\!}}\mathrm{(s)} ~+~ \underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{\phantom{_1\!}\mathrm{F_2}\phantom{_1\!}}\mathrm{(g)} \longrightarrow \underset{\begin{subarray}{c} +4 \\ +4 \end{subarray}}{\phantom{_1\!}\mathrm{Mn}\phantom{_1\!}} \underset{\begin{subarray}{c} -1 \\ -1 \end{subarray}}{\phantom{_1\!}\mathrm{F_4}\phantom{_1\!}}\mathrm{(s)} \end{align*}\]
Mn undergoes oxidation and F2 undergoes reduction.
\[\begin{alignat*}{3} \textrm{Ox. half-reaction} &\quad & \mathrm{Mn} &\; &\longrightarrow &\; \mathrm{Mn^{4+}} + 4~e^- \\ \textrm{Red. half-reaction} &\quad & \mathrm{F_2} + 2~e^- &\; &\longrightarrow &\; 2~\mathrm{F^-}\\[1.0ex] \textit{balance e}^- &\quad & \mathrm{Mn} &\; &\longrightarrow &\; \mathrm{Mn^{4+}} + 4~e^- \\ &\quad & 2~( \mathrm{F_2} + 2~e^- &\; &\longrightarrow &\; 2~\mathrm{F^-})\\[1.0ex] \textit{4 mol e}^- &\quad & \mathrm{Mn} &\; &\longrightarrow &\; \mathrm{Mn^{4+}} + 4~e^- \\ &\quad & 2~\mathrm{F_2} + 4~e^- &\; &\longrightarrow &\; 4~\mathrm{F^-} \\[1.0ex] \textit{combine} &\quad & \mathrm{Mn} + 2~\mathrm{F_2} &\; &\longrightarrow &\; \mathrm{Mn^{4+}} + 4~\mathrm{F^-} \\[1.0ex] \textrm{Balanced reaction} &\quad & \mathrm{Mn(s)} + 2~\mathrm{F_2(g)} &\; &\longrightarrow &\; \mathrm{MnF_4(s)} \end{alignat*}\]
Concept: oxidation state; oxidation-reduction reaction; balancing equations
Determine the half-reactions, the amount (in mol) of electrons transferred, and the overall balanced reaction (with phase labels) for the following redox reaction.
\[\begin{align*} \mathrm{Na(s)} + \mathrm{I_2(g)} \longrightarrow \mathrm{NaI(s)} \end{align*}\]
Solution
Answer:
\[\begin{align*} \underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{\phantom{_1\!}\mathrm{Na}\phantom{_1\!}}\mathrm{(s)} ~+~ \underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{\phantom{_1\!}\mathrm{I_2}\phantom{_1\!}}\mathrm{(g)} \longrightarrow \underset{\begin{subarray}{c} +4 \\ +4 \end{subarray}}{\phantom{_1\!}\mathrm{Na}\phantom{_1\!}} \underset{\begin{subarray}{c} -1 \\ -1 \end{subarray}}{\phantom{_1\!}\mathrm{I_2}\phantom{_1\!}}\mathrm{(s)} \end{align*}\]
Na undergoes oxidation and I2 undergoes reduction.
\[\begin{alignat*}{3} \textrm{Ox. half-reaction} &\quad & \mathrm{Na} &\; &\longrightarrow &\; \mathrm{Na^+} + e^- \\ \textrm{Red. half-reaction} &\quad & \mathrm{I_2} + 2~e^- &\; &\longrightarrow &\; 2~\mathrm{I^-}\\[1.0ex] \textit{balance e}^- &\quad & 2~(\mathrm{Na} &\; &\longrightarrow &\; \mathrm{Na^+} + e^-) \\ &\quad & \mathrm{I_2} + 2~e^- &\; &\longrightarrow &\; 2~\mathrm{I^-}\\[1.0ex] \textit{2 mol e}^- &\quad & 2~\mathrm{Na} &\; &\longrightarrow &\; 2~\mathrm{Na^+} + 2~e^- \\ &\quad & \mathrm{I_2} + 2~e^- &\; &\longrightarrow &\; 2~\mathrm{I^-} \\[1.0ex] \textit{combine} &\quad & 2~\mathrm{Na} + \mathrm{I_2} &\; &\longrightarrow &\; 2~\mathrm{Na^+} + 2~\mathrm{I^-} \\[1.0ex] \textrm{Balanced reaction} &\quad & 2~\mathrm{Na(s)} + \mathrm{I_2(g)} &\; &\longrightarrow &\; 2~\mathrm{NaI(s)} \end{alignat*}\]
Concept: oxidation state; oxidation-reduction reaction; balancing equations
Determine the half-reactions, the amount (in mol) of electrons transferred, and the overall balanced reaction (with phase labels) for the following redox reaction.
\[\begin{align*} \mathrm{Mg(s)} + \mathrm{Cl_2(g)} \longrightarrow \mathrm{MgCl_2(s)} \end{align*}\]
Solution
Answer:
\[\begin{align*} \underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{\phantom{_1\!}\mathrm{Mg}\phantom{_1\!}}\mathrm{(s)} ~+~ \underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{\phantom{_1\!}\mathrm{Cl_2}\phantom{_1\!}}\mathrm{(g)} \longrightarrow \underset{\begin{subarray}{c} +2 \\ +2 \end{subarray}}{\phantom{_1\!}\mathrm{Mg}\phantom{_1\!}} \underset{\begin{subarray}{c} -1 \\ -1 \end{subarray}}{\phantom{_1\!}\mathrm{Cl_2}\phantom{_1\!}}\mathrm{(s)} \end{align*}\]
Mg undergoes oxidation and Cl2 undergoes reduction.
\[\begin{alignat*}{3} \textrm{Ox. half-reaction} &\quad & \mathrm{Mg} &\; &\longrightarrow &\; \mathrm{Mg^+} + 2~e^- \\ \textrm{Red. half-reaction} &\quad & \mathrm{Cl_2} + 2~e^- &\; &\longrightarrow &\; 2~\mathrm{Cl^-}\\[1.0ex] \textit{balance e}^- &\quad & \mathrm{Mg} &\; &\longrightarrow &\; \mathrm{Mg^+} + 2~e^- \\ &\quad & \mathrm{Cl_2} + 2~e^- &\; &\longrightarrow &\; 2~\mathrm{Cl^-}\\[1.0ex] \textit{2 mol e}^- &\quad & \mathrm{Mg} &\; &\longrightarrow &\; \mathrm{Mg^+} + 2~e^- \\ &\quad & \mathrm{Cl_2} + 2~e^- &\; &\longrightarrow &\; 2~\mathrm{Cl^-} \\[1.0ex] \textit{combine} &\quad & \mathrm{Mg} + \mathrm{Cl_2} &\; &\longrightarrow &\; \mathrm{Mg^+} + 2~\mathrm{Cl^-} \\[1.0ex] \textrm{Balanced reaction} &\quad & \mathrm{Mg(s)} + \mathrm{Cl_2(g)} &\; &\longrightarrow &\; \mathrm{MgCl_2(s)} \end{alignat*}\]
Concept: oxidation state; oxidation-reduction reaction; balancing equations
Determine the half-reactions, the amount (in mol) of electrons transferred, and the overall balanced reaction (with phase labels) for the following redox reaction.
\[\begin{align*} \mathrm{Fe(s)} + \mathrm{O_2(g)} \longrightarrow \mathrm{Fe_2O_3(s)} \end{align*}\]
Solution
Answer:
\[\begin{align*} \underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{\phantom{_1\!}\mathrm{Fe}\phantom{_1\!}}\mathrm{(s)} ~+~ \underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{\phantom{_1\!}\mathrm{O_2}\phantom{_1\!}}\mathrm{(g)} \longrightarrow \underset{\begin{subarray}{c} +3 \\ +6 \end{subarray}}{\phantom{_1\!}\mathrm{Fe_2}\phantom{_1\!}} \underset{\begin{subarray}{c} -2 \\ -6 \end{subarray}}{\phantom{_1\!}\mathrm{O_3}\phantom{_1\!}}\mathrm{(s)} \end{align*}\]
Fe undergoes oxidation and O\(_2\) undergoes reduction.
\[\begin{alignat*}{3} \textrm{Ox. half-reaction} &\quad & 2~\mathrm{Fe} &\; &\longrightarrow &\; 2~\mathrm{Fe^{3+}} + 6~e^- \\ \textrm{Red. half-reaction} &\quad & \mathrm{O_2} + 4~e^- &\; &\longrightarrow &\; 2~\mathrm{O^{2-}} \\[1.0ex] \textit{balance e}^- &\quad & 2 ( 2~\mathrm{Fe} &\; &\longrightarrow &\; 2~\mathrm{Fe^{3+}} + 6~e^- ) \\ &\quad & 3 ( \mathrm{O_2} + 4~e^- &\; &\longrightarrow &\; 2~\mathrm{O^{2-}} ) \\[1.0ex] \textit{12 mol e}^- &\quad & 4~\mathrm{Fe} &\; &\longrightarrow &\; 4~\mathrm{Fe^{3+}} + 12~e^- \\ &\quad & 3~\mathrm{O_2} + 12~e^- &\; &\longrightarrow &\; 6~\mathrm{O^{2-}} \\[1.0ex] \textit{combine} &\quad & 4~\mathrm{Fe} + 3~\mathrm{O_2} &\; &\longrightarrow &\; 4~\mathrm{Fe^{3+}} + 6~\mathrm{O^{2-}} \\[1.0ex] \textrm{Balanced reaction} &\quad & 4~\mathrm{Fe(s)} + 3~\mathrm{O_2(g)} &\; &\longrightarrow &\; 2~\mathrm{Fe_2O_3(s)} \end{alignat*}\]
Concept: oxidation state; oxidation-reduction reaction; balancing equations
Determine the half-reactions, the amount (in mol) of electrons transferred, and the overall balanced reaction (with phase labels) for the following redox reaction.
\[\begin{align*} \mathrm{Cr^{3+}(aq)} + \mathrm{Mn^{2+}(aq)} \longrightarrow \mathrm{Mn^{6+}(aq)} + \mathrm{Cr(s)} \end{align*}\]
Solution
Answer:
\[\begin{align*} \underset{\begin{subarray}{c} +3\\ +3 \end{subarray}}{\phantom{_1\!}\mathrm{Cr^{3+}}\phantom{_1\!}}\mathrm{(aq)} ~+~ \underset{\begin{subarray}{c} +2\\ +2 \end{subarray}}{\phantom{_1\!}\mathrm{Mn^{2+}}\phantom{_1\!}}\mathrm{(aq)} \longrightarrow \underset{\begin{subarray}{c} +6 \\ +6 \end{subarray}}{\phantom{_1\!}\mathrm{Mn^{6+}}\phantom{_1\!}}\mathrm{(aq)} \underset{\begin{subarray}{c} 0 \\ 0 \end{subarray}}{\phantom{_1\!}\mathrm{Cr}\phantom{_1\!}}\mathrm{(s)} \end{align*}\]
Mn2+ undergoes oxidation and Cr3+ undergoes reduction.
\[\begin{alignat*}{3} \textrm{Ox. half-reaction} &\quad & \mathrm{Mn^{2+}} &\; &\longrightarrow &\; \mathrm{Mn^{6+}} + 4~e^- \\ \textrm{Red. half-reaction} &\quad & \mathrm{Cr^{3+}} + 3~e^- &\; &\longrightarrow &\; \mathrm{Cr} \\[1.0ex] \textit{balance e}^- &\quad & 3(\mathrm{Mn^{2+}} &\; &\longrightarrow &\; \mathrm{Mn^{6+}} + 4~e^-) \\ &\quad & 4(\mathrm{Cr^{3+}} + 3~e^- &\; &\longrightarrow &\; \mathrm{Cr}) \\[1.0ex] \textit{12 mol e}^- &\quad & 3~\mathrm{Mn^{2+}} &\; &\longrightarrow &\; 3~\mathrm{Mn^{6+}} + 12~e^- \\ &\quad & 4~\mathrm{Cr^{3+}} + 12~e^- &\; &\longrightarrow &\; 4~\mathrm{Cr} \\[1.0ex] \textit{combine} &\quad & 4~\mathrm{Cr^{3+}} + 3~\mathrm{Mn^{2+}} &\; &\longrightarrow &\; 3~\mathrm{Mn^{6+}} + 4~\mathrm{Cr} \\[1.0ex] \textrm{Balanced reaction} &\quad & 4~\mathrm{Cr^{3+}(aq)} + 3~\mathrm{Mn^{2+}(aq)} &\; &\longrightarrow &\; 3~\mathrm{Mn^{6+}(aq)} + 4~\mathrm{Cr(s)} \end{alignat*}\]
Concept: oxidation state; oxidation-reduction reaction; balancing equations
Stoichiometry
Balance the following equation.
\[\begin{align*} \rule[-1.0pt]{2em}{0.5pt} \mathrm{Ag_2O(s)} ~\longrightarrow~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{Ag(s)} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{O_2(g)} \end{align*}\]
Additionally, consider a 4.260 × 103 mg sample of impure silver oxide that, when completely decomposes, yields 283 mg of O2(g). Assuming that the silver oxide is the only source of oxygen, what is the mass percent of silver oxide in the sample?
Solution
Answer: 96.2 %
\[\begin{align*} \mathrm{2~Ag_2O(s)} \longrightarrow \mathrm{4~Ag(s)} + \mathrm{O_2(g)} \end{align*}\]
\[\begin{align*} w(\mathrm{Ag_2O})\% &= w(\mathrm{Ag_2O}) \times 100~\% \\[1.5ex] &= \biggl\{ m(\mathrm{Ag_2O}) ~ m(\mathrm{sample})^{-1} \biggl\} ~ \times 100~\% \\[1.5ex] &= \biggl\{ n(\mathrm{Ag_2O})~ M(\mathrm{Ag_2O})~ m(\mathrm{sample})^{-1} \biggl\}~ \times 100~\% \\[1.5ex] &= \biggl\{ n(\mathrm{O_2})~ r(\mathrm{Ag_2O,O_2}) ~ M(\mathrm{Ag_2O})~ m(\mathrm{sample})^{-1} \biggl\}~ \times 100~\% \\[1.5ex] &= \biggl\{ m(\mathrm{O_2}) ~ M(\mathrm{O_2})^{-1} ~ r(\mathrm{Ag_2O,O_2}) ~ M(\mathrm{Ag_2O})~ m(\mathrm{sample})^{-1} \biggl\}~ \times 100~\% \\[1.5ex] &= \biggl\{ 283~\mathrm{mg~O_2} \left ( \dfrac{\mathrm{g}}{10^3~\mathrm{mg}} \right ) \left ( \dfrac{\mathrm{mol~O_2}}{\mathrm{32.00~g~O_2}} \right ) \left ( \dfrac{2~\mathrm{mol~Ag_2O}}{\mathrm{mol~O_2}} \right ) \left ( \dfrac{231.74~\mathrm{g}}{\mathrm{mol~Ag_2O}} \right ) \\ &\phantom{=~\biggl\{} \biggr[ 4.260\times 10^{3}~\mathrm{mg~sample} \left ( \frac{\mathrm{g}}{10^3~\mathrm{mg}} \right ) \biggr]^{-1} \biggl\}~ \times 100~\% \\[1.5ex] &= \left ( \dfrac{4.0\bar{9}89~\mathrm{g~Ag_2O}} {4.260~\mathrm{g~sample}} \right ) \times 100~\% \\[1.5ex] &= 96.\bar{2}18~\% = 96.2~\% \end{align*}\]
Concept: balancing equations; stoichiometry; percent composition by mass
Balance the following equation.
\[\begin{align*} \rule[-1.0pt]{2em}{0.5pt} \mathrm{NH_3} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{Cl_2} ~\longrightarrow~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{NCl_3} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{HCl} \end{align*}\]
Additionally, what mass (in g) of HCl is produced if 1.27 g of NH3 reacts with 4.53 g of Cl2?
Solution
Answer: 2.33 g
\[\begin{align*} \mathrm{NH_3} + \mathrm{3~Cl_2} \longrightarrow \mathrm{NCl_3} + \mathrm{3~HCl} \end{align*}\]
\[\begin{align*} m(\mathrm{HCl}) &= n(\mathrm{HCl}) ~ M(\mathrm{HCl}) \\[1.5ex] &= m(\mathrm{NH_3}) ~ M(\mathrm{NH_3})^{-1} ~ r(\mathrm{HCl,NH_3}) ~ M(\mathrm{HCl}) \\[1.5ex] &= 1.27~\mathrm{g~NH_3} \left ( \dfrac{\mathrm{mol~NH_3}}{\mathrm{17.04~g}} \right ) \left ( \dfrac{3~\mathrm{mol~HCl}}{\mathrm{mol~NH_3}} \right ) \left ( \dfrac{36.46~\mathrm{g}}{\mathrm{mol~HCl}} \right ) \\[1.5ex] &= 8.1\bar{5}21~\mathrm{g} = 8.15~\mathrm{g} \\[3ex] m(\mathrm{HCl}) &= n(\mathrm{HCl}) ~ M(\mathrm{HCl}) \\[1.5ex] &= m(\mathrm{Cl_2}) ~ M(\mathrm{Cl_2})^{-1} ~ r(\mathrm{HCl,Cl_2}) ~ M(\mathrm{HCl}) \\[1.5ex] &= 4.53~\mathrm{g~Cl_2} \left ( \dfrac{\mathrm{mol~Cl_2}}{\mathrm{70.90~g}} \right ) \left ( \dfrac{3~\mathrm{mol~HCl}}{\mathrm{3~mol~Cl_2}} \right ) \left ( \dfrac{36.46~\mathrm{g}}{\mathrm{mol~HCl}} \right ) \\[1.5ex] &= 2.3\bar{2}95~\mathrm{g} = 2.33~\mathrm{g} \end{align*}\]
Cl2 is the limiting reactant. 2.33 g of HCl is produced.
Concept: balancing equations; stoichiometry; limiting reactant
Balance the following equation.
\[\begin{align*} \rule[-1.0pt]{2em}{0.5pt} \mathrm{C_3H_6} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{NH_3} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{O_2} ~\longrightarrow~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{C_3NH_3} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{H_2O} \end{align*}\]
Additionally, determine the following:
- What is the limiting reactant if 4.25 g of C3H6 reacts with 3.14 g of NH3 and 6.12 g of O2?
- What mass (in g) of C3NH3 can theoretically be produced given the information in (a)?
- What mass (in g) of C3NH3, NH3, and O2 would theoretically be leftover given the information in (a)?
- If 1.94 g of C3NH3 was produced in an experiment, what is the percent yield?
Solution
Answer:
\[\begin{align*} \mathrm{2~C_3H_6} + \mathrm{2~NH_3} + \mathrm{3~O_2} \longrightarrow \mathrm{2~C_3NH_3} + \mathrm{6~H_2O} \end{align*}\]
A
\[\begin{align*} n(\mathrm{C_3NH_3}) &= m(\mathrm{C_3H_6}) ~ M(\mathrm{C_3H_6})^{-1} ~ r(\mathrm{C_3NH_3,C_3H_6}) ~ \\[1.5ex] &= 4.25~\mathrm{g~C_3H_6} \left ( \dfrac{\mathrm{mol~C_3H_6}}{\mathrm{42.09~g}} \right ) \left ( \dfrac{2~\mathrm{mol~C_3NH_3}}{2~\mathrm{mol~C_3H_6}} \right )\\[1.5ex] &= 0.10\bar{0}97~\mathrm{mol} \\[1.5ex] n(\mathrm{C_3NH_3}) &= m(\mathrm{NH_3}) ~ M(\mathrm{NH_3})^{-1} ~ r(\mathrm{C_3NH_3,NH_3}) ~ \\[1.5ex] &= 3.14~\mathrm{g~NH_3} \left ( \dfrac{\mathrm{mol~NH_3}}{\mathrm{17.04~g}} \right ) \left ( \dfrac{2~\mathrm{mol~C_3NH_3}}{2~\mathrm{mol~NH_3}} \right )\\[1.5ex] &= 0.18\bar{4}27~\mathrm{mol} \\[3ex] n(\mathrm{C_3NH_3}) &= m(\mathrm{O_2}) ~ M(\mathrm{O_2})^{-1} ~ r(\mathrm{C_3NH_3,O_2}) ~ \\[1.5ex] &= 6.12~\mathrm{g~O_2} \left ( \dfrac{\mathrm{mol~O_2}}{\mathrm{32.00~g}} \right ) \left ( \dfrac{2~\mathrm{mol~C_3NH_3}}{3~\mathrm{mol~O_2}} \right ) \\[1.5ex] &= 0.12\bar{7}50~\mathrm{mol} \end{align*}\]
C3H6 is limiting.
B
\[\begin{align*} m(\mathrm{C_3NH_3}) &= 0.10\bar{0}97~\mathrm{mol} \left ( \dfrac{53.06~\mathrm{g}}{\mathrm{mol~C_3NH_3}} \right ) \\[1.5ex] &= 5.3\bar{5}76~\mathrm{g} = 5.36~\mathrm{g} \end{align*}\]
5.36 g of C3NH3 can be produced.
C
\[\begin{align*} m(\mathrm{C_3H_6})_{\mathrm{leftover}} &= m(\mathrm{C_3H_6})_{\mathrm{initial}} - m(\mathrm{C_3H_6})_{\mathrm{reacted}} \\[1.5ex] &= m(\mathrm{C_3H_6})_{\mathrm{initial}} - \left [ n(\mathrm{C_3NH_3}) ~ r(\mathrm{C_3H_6,C_3NH_3}) ~ M(\mathrm{C_3H_6}) \right ] \\[1.5ex] &= 4.25~\mathrm{g~C_3H_6} - \left [ 0.10\bar{0}97~\mathrm{mol~C_3NH_3} \left ( \dfrac{\mathrm{2~mol~NH_3}}{2~\mathrm{mol~C_3NH_3}} \right ) \left ( \dfrac{42.09~\mathrm{g}}{\mathrm{mol~C_3H_6}} \right ) \right ] \\[1.5ex] &= 4.25~\mathrm{g} - 4.2\bar{4}98~\mathrm{g} \\[1.5ex] &= 0~\mathrm{g}\\[3ex] m(\mathrm{NH_3})_{\mathrm{leftover}} &= m(\mathrm{NH_3})_{\mathrm{initial}} - m(\mathrm{NH_3})_{\mathrm{reacted}} \\[1.5ex] &= m(\mathrm{NH_3})_{\mathrm{initial}} - \left [ n(\mathrm{C_3NH_3}) ~ r(\mathrm{NH_3,C_3NH_3}) ~ M(\mathrm{NH_3}) \right ] \\[1.5ex] &= 3.14~\mathrm{g~NH_3} - \left [ 0.10\bar{0}97~\mathrm{mol~C_3NH_3} \left ( \dfrac{\mathrm{2~mol~NH_3}}{2~\mathrm{mol~C_3NH_3}} \right ) \left ( \dfrac{17.04~\mathrm{g}}{\mathrm{mol~NH_3}} \right ) \right ] \\[1.5ex] &= 3.14~\mathrm{g} - 1.7\bar{2}05~\mathrm{g} \\[1.5ex] &= 1.4\bar{1}94~\mathrm{g} = 1.42~\mathrm{g}\\[3ex] m(\mathrm{O_2})_{\mathrm{leftover}} &= m(\mathrm{O_2})_{\mathrm{initial}} - m(\mathrm{O_2})_{\mathrm{reacted}} \\[1.5ex] &= m(\mathrm{O_2})_{\mathrm{initial}} - \left [ n(\mathrm{C_3NH_3}) ~ r(\mathrm{O_2,C_3NH_3}) ~ M(\mathrm{O_2}) \right ] \\[1.5ex] &= 6.12~\mathrm{g~O_2} - \left [ 0.10\bar{0}97~\mathrm{mol~C_3NH_3} \left ( \dfrac{\mathrm{3~mol~O_2}}{2~\mathrm{mol~C_3NH_3}} \right ) \left ( \dfrac{32.00~\mathrm{g}}{\mathrm{mol~O_2}} \right ) \right ] \\[1.5ex] &= 6.12~\mathrm{g} - 4.8\bar{4}65~\mathrm{g}\\[1.5ex] &= 1.2\bar{7}34~\mathrm{g} = 1.27~\mathrm{g} \end{align*}\]
D
\[\begin{align*} \%~\mathrm{yield} &= \left ( \dfrac{m(\mathrm{actual})}{m(\mathrm{theoretical})} \right ) 100~\% \\[1.5ex] &= \left ( \dfrac{1.94~\mathrm{g}}{5.3\bar{5}86~\mathrm{g}}\right ) 100~\% \\[1.5ex] &= 36.\bar{2}03~\% = 36.2~\% \end{align*}\]
Concept: balancing equations; limiting reactant; percent yield; stoichiometry
Balance the following equation.
\[\begin{align*} \rule[-1.0pt]{2em}{0.5pt} \mathrm{C_6H_5NO_2} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{C_6H_{14}O_4} ~\overset{\mathrm{Zn}}{ \underset{\mathrm{KOH}}{\longrightarrow}}~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{(C_6H_5N)_2} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{C_6H_{12}O_4} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{H_2O} \end{align*}\]
Additionally, determine the following:
- What volume (in mL) of C6H5NO2 (ρ = 1.20 g mL–1) must be allowed to react with an excess of C6H14O4 to produce 6.21 g of (C6H5N)2 if the percent yield is 83.7 %?
- If 0.17 L of C6H5NO2 (ρ = 1.20 g mL–1) and 0.52 L C6H14O4 (ρ = 1.12 g mL–1) react to yield 64.4 g of (C6H5N)2, what is the limiting reactant and what is the percent yield of the reaction?
Solution
Answer: 8.35 mL; C6H5NO2 is limiting; 43 %
\[\begin{align*} \mathrm{2~C_3H_6} + \mathrm{2~NH_3} + \mathrm{3~O_2} \longrightarrow \mathrm{2~C_3NH_3} + \mathrm{6~H_2O} \end{align*}\]
A
\[\begin{align*} V(\mathrm{C_6H_5NO_2}) &= m(\mathrm{C_6H_5NO_2}) ~ \rho (\mathrm{C_6H_5NO_2})^{-1} \\[1.5ex] &= m[\mathrm{(C_6H_5N)_2}] ~ M[\mathrm{(C_6H_5N)_2}]^{-1} ~ r[\mathrm{C_6H_5NO_2,(C_6H_5N)_2}] ~ M(\mathrm{C_6H_5NO_2}) ~ \rho(\mathrm{C_6H_5NO_2})^{-1}\\[1.5ex] &= 7.42~\mathrm{g~(C_6H_5N)_2} \left ( \dfrac{\mathrm{mol~(C_6H_5N)_2}}{182.24~\mathrm{g}} \right ) \left ( \dfrac{2~\mathrm{mol~C_6H_5NO_2}}{\mathrm{mol~(C_6H_5N)_2}} \right ) \left ( \dfrac{123.12~\mathrm{g}}{\mathrm{mol~C_6H_5NO_2}} \right ) \\ &\phantom{=}~ \left ( \dfrac{\mathrm{mL~}}{1.20~\mathrm{g~C_6H_5NO_2}} \right ) \\[1.5ex] &= 8.3\bar{5}48~\mathrm{mL} \\[1.5ex] &= 8.35~\mathrm{mL} \end{align*}\]
B
\[\begin{align*} m[\mathrm{(C_6H_5N)_2}] &= n[\mathrm{(C_6H_5N)_2}] ~ M[\mathrm{(C_6H_5N)_2}]\\[1.5ex] &= V(\mathrm{C_6H_5NO_2}) ~ \rho(\mathrm{C_6H_5NO_2}) ~ M(\mathrm{C_6H_5NO_2})^{-1} ~ r[\mathrm{(C_6H_5N)_2,C_6H_5NO)_2}] ~ M[\mathrm{(C_6H_5N)_2}] \\[1.5ex] &= 0.17~\mathrm{L~C_6H_5NO_2} \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \left ( \dfrac{1.20~\mathrm{g~C_6H_5NO_2}}{\mathrm{mL}} \right ) \left ( \dfrac{\mathrm{mol~C_6H_5NO_2}}{123.12~\mathrm{g}} \right ) \\ &\phantom{=}~ \left ( \dfrac{\mathrm{mol~(C_6H_5N)_2}}{2~\mathrm{mol~C_6H_5NO_2}} \right ) \left ( \dfrac{182.24~\mathrm{g}}{\mathrm{mol~(C_6H_5N)_2}} \right ) \\[1.5ex] &= 1\bar{5}0.9~\mathrm{g} = 150~\mathrm{g}\\[1.5ex] m[\mathrm{(C_6H_5N)_2}] &= n[\mathrm{(C_6H_5N)_2}] ~ M[\mathrm{(C_6H_5N)_2}]\\[1.5ex] &= V(\mathrm{C_6H_{14}O_4}) ~ \rho(\mathrm{C_6H_{14}O_4}) ~ M(\mathrm{C_6H_{14}O_4})^{-1} ~ r[\mathrm{(C_6H_5N)_2,C_6H_{14}O_4}] ~ M[\mathrm{(C_6H_5N)_2}] \\[1.5ex] &= 0.52~\mathrm{L~C_6H_{14}O_4} \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \left ( \dfrac{1.12~\mathrm{g~C_6H_{14}O_4}}{\mathrm{mL}} \right ) \left ( \dfrac{\mathrm{mol~C_6H_{14}O_4}}{150.20~\mathrm{g}} \right ) \\ &\phantom{=}~ \left ( \dfrac{\mathrm{mol~(C_6H_5N)_2}}{4~\mathrm{mol~C_6H_{14}O_4}} \right ) \left ( \dfrac{182.24~\mathrm{g}}{\mathrm{mol~(C_6H_5N)_2}} \right ) \\[1.5ex] &= 1\bar{7}6.6~\mathrm{g} = 180~\mathrm{g} \end{align*}\]
C
\[\begin{align*} \%~\mathrm{yield} &= \left ( \dfrac{m[\mathrm{(C_6H_5N)_2}]_{\mathrm{actual}}}{m[\mathrm{(C_6H_5N)_2}]_{\mathrm{theoretical}}} \right ) 100~\% \\[1.5ex] &= \left ( \dfrac{64.4~\mathrm{g}}{1\bar{5}0.9~\mathrm{g}}\right ) 100~\% = 4\bar{2}.67~\% = 43~\% \end{align*}\]
Concept: balancing equations; limiting reactant; percent yield; stoichiometry
Balance the following equation.
\[\begin{align*} \rule[-1.0pt]{2em}{0.5pt} \mathrm{Fe} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{Br_2} ~\longrightarrow~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{NaBr} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{CO_2} ~+~ \rule[-1.0pt]{2em}{0.5pt} \mathrm{Fe_3O_4} \end{align*}\]
Additionally, an alloy contains 84.2 % Fe and 15.8 % Ni (by mass). A 6.41 g sample of the alloy reacts with 8.30 g Na2CO3 with excess Br2.
- What mass (in g) of Fe3O4 can be produced?
- What mass (in g) of Fe3O4 is produced if the percent yield of the reaction is 94.3 %?
Solution
Answer: 4.53 g; 4.27 g
\[\begin{align*} \mathrm{3~Fe} + \mathrm{4~Br_2} \longrightarrow \mathrm{8~NaBr} + \mathrm{4~CO_2} + \mathrm{Fe_3O_4} \end{align*}\]
A
\[\begin{align*} m(\mathrm{Fe_3O_4}) &= n(\mathrm{Fe_3O_4}) ~ M(\mathrm{Fe_3O_4}) \\[1.5ex] &= m(\mathrm{sample}) ~ w(\mathrm{sample}) ~ M(\mathrm{Fe})^{-1} ~ r(\mathrm{Fe_3O_4,Fe}) ~ M(\mathrm{Fe_3O_4}) \\[1.5ex] &= 6.41~\mathrm{g~sample} \left ( \dfrac{84.2~\%~\mathrm{Fe}}{100~\%~\mathrm{sample}} \right ) \left ( \dfrac{\mathrm{mol~Fe}}{55.85~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{mol~Fe_3O_4}}{\mathrm{3~mol~Fe}} \right ) \left ( \dfrac{231.55~\mathrm{g}}{\mathrm{mol~Fe_3O_4}} \right ) \\[1.5ex] &= 7.4\bar{5}88~\mathrm{g} = 7.46~\mathrm{g}\\[1.5ex] m(\mathrm{Fe_3O_4}) &= n(\mathrm{Fe_3O_4}) ~ M(\mathrm{Fe_3O_4}) \\[1.5ex] &= m(\mathrm{Na_2CO_3}) ~ M(\mathrm{Na_2CO_3})^{-1} ~ r(\mathrm{Fe_3O_4,Na_2CO_3}) ~ M(\mathrm{Fe_3O_4}) \\[1.5ex] &= 8.30~\mathrm{g~Na_2CO_3} \left ( \dfrac{\mathrm{mol~Na_2CO_3}}{105.99~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{mol~Fe_3O_4}}{\mathrm{4~mol~Fe}} \right ) \left ( \dfrac{231.55~\mathrm{g}}{\mathrm{mol~Fe_3O_4}} \right ) \\[1.5ex] &= 4.5\bar{3}31~\mathrm{g} \\[1.5ex] &= 4.53~\mathrm{g} \end{align*}\]
B
\[\begin{align*} \%~\mathrm{yield} &= \left ( \dfrac{m(\mathrm{Fe_3O_4})_{\mathrm{actual}}}{m(\mathrm{Fe_3O_4})_{\mathrm{theoretical}}} \right ) 100~\% \longrightarrow \\[1.5ex] m(\mathrm{Fe_3O_4})_{\mathrm{actual}} &= \left ( \dfrac{\mathrm{\%~yield}}{100~\%} \right ) m(\mathrm{Fe_3O_4})_{\mathrm{theoretical}}\\[1.5ex] &= \left ( \dfrac{94.3~\%}{100~\%} \right ) 4.5\bar{3}31~\mathrm{g} \\[1.5ex] &= 4.2\bar{7}47~\mathrm{g} = 4.27~\mathrm{g} \end{align*}\]
Concept: balancing equations; limiting reactant; percent yield; stoichiometry
A mixture contained no fluorine compound except methyl fluoroacetate, FCH2COOCH3 (M(FCH2COOCH3) = 92.08 g mol–1). When chemically treated, all the fluorine was converted to CaF2 (M(CaF2) = 78.08 g mol–1). The mass of CaF2 obtained was 20.1 g. Find the mass (in g) of methyl fluoroacetate in the original mixture.
Solution
Answer: 47.4 g
\[\begin{align*} m(\mathrm{FCH_2COOCH_3}) &= n(\mathrm{FCH_2COOCH_3}) ~ M(\mathrm{FCH_2COOCH_3}) \\[1.5ex] &= m(\mathrm{CaF_2}) ~ M(\mathrm{CaF_2})^{-1} ~ r(\mathrm{F^-,CaF_2}) ~ r(\mathrm{FCH_2COOCH_3,F^-})~ M(\mathrm{FCH_2COOCH_3}) \\[1.5ex] &= 20.1~\mathrm{g~CaF_2} \left ( \dfrac{\mathrm{mol~CaF_2}}{78.08~\mathrm{g}} \right ) \left ( \dfrac{2~\mathrm{mol~F^-}}{\mathrm{mol~CaF_2}} \right ) \left ( \dfrac{\mathrm{mol~FCH_2COOCH_3}}{\mathrm{mol~F^-}} \right )\\ &\phantom{=}~ \left ( \dfrac{92.08~\mathrm{g}}{\mathrm{mol~FCH_2COOCH_3}} \right )\\[1.5ex] &= 47.\bar{4}07~\mathrm{g} = 47.4~\mathrm{g} \end{align*}\]
Concept: stoichiometry
A 1.62 g sample of a metal chloride, MCl2, is dissolved in water and treated with excess aqueous silver nitrate. The silver chloride that formed weighed 3.46 g. Calculate the molar mass (in g mol–1) of the metal, M, and identify the metal.
Solution
Answer: 63.3 g mol–1; Cu
\[\begin{align*} \mathrm{MCl_2(aq)} + \mathrm{2~AgNO_3(aq)} \longrightarrow \mathrm{2~AgCl(s)} + \mathrm{M(NO_3)_2(aq)} \end{align*}\]
\[\begin{align*} M(\mathrm{M}) &= m(\mathrm{M}) ~ n(\mathrm{M})^{-1} \\[1.5ex] &= \biggl\{ m(\mathrm{MCl_2}) ~ n(\mathrm{MCl_2})^{-1} \biggl\} ~ - ~ 2~M(\mathrm{Cl}) \\[1.5ex] &= \biggl\{ m(\mathrm{MCl_2}) \biggr[ m(\mathrm{AgCl})~ M(\mathrm{AgCl})^{-1}~ r(\mathrm{Cl^-,AgCl})~ r(\mathrm{MCl_2,Cl^-}) \biggr]^{-1} \biggl\} ~ - ~ \! 2~M(\mathrm{Cl}) \\ &= \biggl\{ 1.62~\mathrm{g~MCl_2} \biggr[ 3.46~\mathrm{g~AgCl} \left ( \dfrac{\mathrm{mol~AgCl}}{143.32~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{mol~Cl^-}}{\mathrm{mol~AgCl}} \right ) \left ( \dfrac{\mathrm{mol~MCl_2}}{\mathrm{2~mol~Cl^-}} \right ) \biggr]^{-1} \biggl\} ~ - ~ \\[1.5ex] &\phantom{=~\biggl\{} \biggr[ 2 ~ \left ( \dfrac{35.45~\mathrm{g}}{\mathrm{mol~Cl}}\right ) \biggr]\\[1.5ex] &= \biggl\{ \dfrac{1.62~\mathrm{g~MCl_2}}{0.012\bar{0}70~\mathrm{mol~MCl_2}} \biggl\} ~ - ~ \biggr[ 2 ~ \left ( \dfrac{35.45~\mathrm{g}}{\mathrm{mol~Cl}}\right ) \biggr]\\[1.5ex] &= 63.\bar{3}07~\mathrm{g} = 63.3~\mathrm{g} \end{align*}\]
Concept: stoichiometry
Which solution has the greatest molar concentration of SO42–?
- 0.060 M H2SO4
- 0.27 M MgSO4
- 0.17 M Na2SO4
- 0.098 M Al2(SO4)3
- 0.22 M CuSO4
Solution
Answer: D
- 0.06 M H2SO4
- 0.27 M MgSO4
- 0.17 M Na2SO4
- 0.29 M Al2(SO4)3
- 0.22 M CuSO4
Concept: electrolytes; molarity
A solution is prepared by dissolving 4.25 g NaCl, 0.175 g KCl, and 0.183 g CaCl2 in water. The volume of the solution is 500.0 mL. What is the molar concentration (in mol L–1) of Cl– in the solution?
Solution
Answer: 0.157 M
NaCl
\[\begin{align*} n(\mathrm{Cl^-}) &= n(\mathrm{NaCl}) ~ r(\mathrm{Cl^-,NaCl}) \\[1.5ex] &= m(\mathrm{NaCl}) ~ M(\mathrm{NaCl})^{-1} ~ r(\mathrm{Cl^-,NaCl}) \\[1.5ex] &= 4.25~\mathrm{g} \left ( \dfrac{\mathrm{mol~NaCl}}{58.44~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{mol~Cl^-}}{\mathrm{mol~NaCl}} \right ) \\[1.5ex] &= 0.072\bar{7}24~\mathrm{mol}\\[3ex] \end{align*}\]
KCl
\[\begin{align*} n(\mathrm{Cl^-}) &= n(\mathrm{KCl}) ~ r(\mathrm{Cl^-,KCl}) \\[1.5ex] &= m(\mathrm{KCl}) ~ M(\mathrm{KCl})^{-1} ~ r(\mathrm{Cl^-,KCl}) \\[1.5ex] &= 0.175~\mathrm{g} \left ( \dfrac{\mathrm{mol~KCl}}{74.55~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{mol~Cl^-}}{\mathrm{mol~KCl}} \right ) \\[1.5ex] &= 0.0023\bar{4}74~\mathrm{mol} \\[3ex] \end{align*}\]
CaCl2
\[\begin{align*} n(\mathrm{Cl^-}) &= n(\mathrm{CaCl_2}) ~ r(\mathrm{Cl^-,CaCl_2}) \\[1.5ex] &= m(\mathrm{CaCl_2}) ~ M(\mathrm{CaCl_2})^{-1} ~ r(\mathrm{Cl^-,CaCl_2}) \\[1.5ex] &= 0.183~\mathrm{g} \left ( \dfrac{\mathrm{mol~CaCl_2}}{110.98~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{2~mol~Cl^-}}{\mathrm{mol~CaCl_2}} \right ) \\[1.5ex] &= 0.0032\bar{9}78~\mathrm{mol} \\[3ex] \end{align*}\]
Cl– molar concentration
\[\begin{align*} c(\mathrm{Cl^-}) &= n(\mathrm{Cl^-}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggr[ (0.072\bar{7}24 + 0.0023\bar{4}74 + 0.0032\bar{9}78)~\mathrm{mol} \biggr] \biggr[ 500.0~\mathrm{mL} \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \biggr]^{-1} \\[1.5ex] &= 0.15\bar{6}73~\mathrm{mol~L^{-1}} = 0.157~\mathrm{mol~L^{-1}} \end{align*}\]
Concept: electrolytes; molarity
In the following reaction, 55.2 mL of potassium sulfate solution was added to excess lead acetate. What is the concentration of K+ in the potassium sulfate solution if 1.23 g of PbSO4 was produced?
\[\begin{align*} \mathrm{K_2SO_4(aq)} + \mathrm{Pb(C_2H_3O_2)_2(aq)} \longrightarrow \mathrm{2~KC_2H_3O_2(aq)} + \mathrm{PbSO_4(s)} \end{align*}\]
Solution
Answer: 0.147 M
\[\begin{align*} c(\mathrm{K^+}) &= n(\mathrm{K^+}) ~ V(\mathrm{solution})^{-1}\\[1.5ex] &= n(\mathrm{K_2SO_4}) ~ r(\mathrm{K^+, K_2SO_4}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= n(\mathrm{PbSO_4}) ~ r(\mathrm{K_2SO_4,PbSO_4}) ~ r(\mathrm{K^+, K_2SO_4}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= m(\mathrm{PbSO_4}) ~ M(\mathrm{PbSO_4})^{-1} ~ r(\mathrm{K_2SO_4,PbSO_4}) ~ r(\mathrm{K^+, K_2SO_4}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= 1.23~\mathrm{g~PbSO_4} \left ( \dfrac{\mathrm{mol~PbSO_4}}{303.26~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{mol~K_2SO_4}}{\mathrm{mol~PbSO_4}} \right ) \left ( \dfrac{2~\mathrm{mol~K^+}}{\mathrm{mol~K_2SO_4}} \right ) \left ( \dfrac{1}{52.2~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 0.15\bar{5}39~\mathrm{mol~L^{-1}} = 0.155~\mathrm{mol~L^{-1}} \end{align*}\]
Concept: electrolytes; molarity; stoichiometry
You mix 285.0 mL of 1.20 M aqueous lead(II) nitrate with 300.0 mL of 1.55 M aqueous potassium iodide. Determine the following.
- the molecular equation for this reaction
- the limiting reactant
- the final molar concentration (in mol L–1) of Pb2+
- the mass (in g) of lead(II) iodide formed
- the final molar concentration (in mol L–1) of K+
- the final molar concentration (in mol L–1) of NO3–
Solution
Answer:
A
\[\begin{align*} \mathrm{Pb(NO_3)_2(aq)} + \mathrm{2~KI(aq)} \longrightarrow \mathrm{PbI_2(s)} + \mathrm{2~KNO_3(aq)}\\ \end{align*}\]
B
\[\begin{align*} c(\mathrm{PbI_2}) &= n(\mathrm{PbI_2}) ~ V(\mathrm{solution})^{-1}\\[1.5ex] &= V(\mathrm{Pb(NO_3)_2}) ~ c(\mathrm{Pb(NO_3)_2}) ~ r(\mathrm{PbI_2,Pb(NO_3)_2}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= 285.0~\mathrm{mL~Pb(NO_3)_2} \left ( \dfrac{1.20~\mathrm{mol~Pb(NO_3)_2}}{\mathrm{L}} \right ) \left ( \dfrac{\mathrm{mol~PbI_2}}{\mathrm{mol~Pb(NO_3)_2}} \right ) \left ( \dfrac{1}{(285.0 + 300.0)~\mathrm{mL}} \right ) \\[1.5ex] &= 0.58\bar{4}61~\mathrm{mol~L^{-1}} \\[3ex] c(\mathrm{PbI_2}) &= n(\mathrm{PbI_2}) ~ V(\mathrm{solution})^{-1}\\[1.5ex] &= V(\mathrm{KI}) ~ c(\mathrm{KI}) ~ r(\mathrm{PbI_2,KI}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= 300.0~\mathrm{mL~KI} \left ( \dfrac{1.55~\mathrm{mol~KI}}{\mathrm{L}} \right ) \left ( \dfrac{\mathrm{mol~PbI_2}}{2~\mathrm{mol~KI}} \right ) \left ( \dfrac{1}{(285.0 + 300.0)~\mathrm{mL}} \right ) \\[1.5ex] &= 0.39\bar{7}43~\mathrm{mol~L^{-1}} \end{align*}\]
C
\[\begin{align*} c(\mathrm{Pb^{2+}})_{\mathrm{final}} &= n(\mathrm{Pb^{2+}}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ n(\mathrm{Pb(NO_3)_2})_{\mathrm{initial}} - n(\mathrm{Pb(NO_3)_2})_{\mathrm{reacted}} \biggl\} ~ r(\mathrm{Pb^{2+},Pb(NO_3)_2}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \Biggl\{ \biggr[ c(\mathrm{Pb(NO_3)_2}) ~ V(\mathrm{Pb(NO_3)_2}) \biggr] - \biggr[ c(\mathrm{PbI_2}) ~ V(\mathrm{solution}) ~ r(\mathrm{PbI_2,Pb(NO_3)_2}) ~ \biggr] \Biggl\} \\ &\phantom{=}~~~ r(\mathrm{Pb^{2+},Pb(NO_3)_2}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \Biggl\{\biggr[ \left ( \dfrac{1.20~\mathrm{mol~Pb(NO_3)_2}}{\mathrm{L}} \right ) \left ( \dfrac{285.0~\mathrm{mL~Pb(NO_3)_2}}{} \right ) \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \biggr] - \\ &\phantom{=}~~~ \biggr[ \left ( \dfrac{0.39\bar{7}43~\mathrm{mol~PbI_2}}{\mathrm{L}} \right ) \left ( \dfrac{(285.0~\mathrm{mL} + 300.0)~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \left ( \dfrac{\mathrm{mol~Pb(NO_3)_2}}{\mathrm{mol~PbI_2}} \right ) \biggr]\Biggl\}\\ &\phantom{=}~ \left ( \dfrac{\mathrm{mol~Pb^{2+}}}{\mathrm{mol~Pb(NO_3)_2}} \right ) \left ( \dfrac{1}{(285.0 + 300.0) ~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 0.18\bar{7}18~\mathrm{mol~L^{-1}} = 0.187~\mathrm{mol~L^{-1}} \end{align*}\]
D
\[\begin{align*} m(\mathrm{PbI_2}) &= c(\mathrm{PbI_2}) ~ V(\mathrm{solution}) ~ M(\mathrm{PbI_2}) \\[1.5ex] &= \dfrac{0.39\bar{7}43~\mathrm{mol~PbI_2}}{\mathrm{L}} \left ( \dfrac{(285.0 + 300.0)~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \left ( \dfrac{461~\mathrm{g}}{\mathrm{mol~PbI_2}} \right ) \\[1.5ex] &= 10\bar{7}.18~\mathrm{g} = 111~\mathrm{g} \end{align*}\]
E
\[\begin{align*} c(\mathrm{K^+}) &= n(\mathrm{K^+}) ~ V(\mathrm{solution})^{-1}\\[1.5ex] &= n(\textrm{KI}) ~ r(\mathrm{K^+,\textrm{KI}})~ V(\mathrm{solution})^{-1} \\[1.5ex] &= c(\textrm{KI}) ~ V(\textrm{KI}) ~ r(\mathrm{K^+,\textrm{KI}})~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \dfrac{1.55~\mathrm{mol~KI}}{\mathrm{L}} \left ( \dfrac{300.0~\mathrm{mL}}{} \right ) \left ( \dfrac{2~\mathrm{mol~K^+}}{\mathrm{2~mol~KNO_3}} \right ) \left ( \dfrac{1}{(285.0 + 300.0)~\mathrm{mL}} \right ) \\[1.5ex] &= 0.79\bar{4}87~\mathrm{mol~L^{-1}} = 0.795~\mathrm{mol~L^{-1}} \end{align*}\]
F
\[\begin{align*} c(\mathrm{NO_3^-}) &= n(\mathrm{NO_3^-}) ~ V(\mathrm{solution})^{-1}\\[1.5ex] &= c(\mathrm{Pb(NO_3)_2}) ~ V(\mathrm{Pb(NO_3)_2}) ~ r(\mathrm{NO_3^{2-},Pb(NO_3)_2})~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \dfrac{1.20~\mathrm{mol~Pb(NO_3)_2}}{\mathrm{L}} \left ( \dfrac{285.0~\mathrm{mL}}{} \right ) \left ( \dfrac{2~\mathrm{mol~NO_3^{2-}}}{\mathrm{mol~Pb(NO_3)_2}} \right ) \left ( \dfrac{1}{(285.0 + 300.0)~\mathrm{mL}} \right ) \\[1.5ex] &= 1.1\bar{6}92~\mathrm{mol~L^{-1}} = 1.17~\mathrm{mol~L^{-1}} \end{align*}\]
Concept: solubility rules; stoichiometry; molarity; limiting reactant
If all of the chloride in a 4.106 g sample of an unknown metal chloride is precipitated as AgCl with 70.90 mL of 0.2010 M AgNO3, what is the percentage of chloride in the sample?
Solution
Answer: 12.30 %
\[\begin{align*} w(\mathrm{Cl^-})\% &= w(\mathrm{Cl^-}) \times 100~\%\\[1.5ex] &= m(\mathrm{Cl^-}) ~ m(\mathrm{sample})^{-1} \times 100~\%\\[1.5ex] &= n(\mathrm{Cl^-}) ~ M(\mathrm{Cl^-}) ~ m(\mathrm{sample})^{-1} \times 100~\%\\[1.5ex] &= n(\mathrm{AgNO_3}) ~ r(\mathrm{Ag^+,AgNO_3}) ~ r(\mathrm{AgCl,Ag^+}) ~ r(\mathrm{Cl^-,Ag^+}) ~ M(\mathrm{Cl^-}) ~ m(\mathrm{sample})^{-1} \times 100~\%\\[1.5ex] &= c(\mathrm{AgNO_3}) ~ V(\mathrm{AgNO_3}) ~ r(\mathrm{Ag^+,AgNO_3}) ~ r(\mathrm{AgCl,Ag^+}) ~ r(\mathrm{Cl^-,Ag^+}) \\ &\phantom{=}~~~ M(\mathrm{Cl^-}) m(\mathrm{sample})^{-1} \times 100~\%\\[1.5ex] &= \dfrac{0.2010~\mathrm{mol~AgNO_3}}{\mathrm{L}} \left ( \dfrac{70.90~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \left ( \dfrac{\mathrm{mol~Ag^+}}{\mathrm{mol~AgNO_3}} \right ) \left ( \dfrac{\mathrm{mol~AgCl}}{\mathrm{mol~Ag^+}} \right ) \left ( \dfrac{\mathrm{mol~Cl^-}}{\mathrm{mol~AgCl}} \right )\\ &\phantom{=}~~ \left ( \dfrac{35.45~\mathrm{g}}{\mathrm{mol~Cl^-}} \right ) \left ( \dfrac{1}{4.106~\mathrm{g}} \right ) \times 100~\% \\[1.5ex] &= 12.3\bar{0}38~\% = 12.30~\% \end{align*}\]
Concept: stoichiometry; molarity; percent composition by mass
If all of the chloride in a 4.106 g sample of an unknown metal chloride is precipitated as AgCl with 70.90 mL of 0.2010 M AgNO3, what is the percentage of chloride in the sample?
Solution
Answer: 12.30 %
\[\begin{align*} w(\mathrm{Cl^-})\% &= w(\mathrm{Cl^-}) \times 100~\%\\[1.5ex] &= m(\mathrm{Cl^-}) ~ m(\mathrm{sample})^{-1} \times 100~\%\\[1.5ex] &= n(\mathrm{Cl^-}) ~ M(\mathrm{Cl^-}) ~ m(\mathrm{sample})^{-1} \times 100~\%\\[1.5ex] &= n(\mathrm{AgNO_3}) ~ r(\mathrm{Ag^+,AgNO_3}) ~ r(\mathrm{AgCl,Ag^+}) ~ r(\mathrm{Cl^-,Ag^+}) ~ M(\mathrm{Cl^-}) ~ m(\mathrm{sample})^{-1} \times 100~\%\\[1.5ex] &= c(\mathrm{AgNO_3}) ~ V(\mathrm{AgNO_3}) ~ r(\mathrm{Ag^+,AgNO_3}) ~ r(\mathrm{AgCl,Ag^+}) ~ r(\mathrm{Cl^-,Ag^+}) \\ &\phantom{=}~~~ M(\mathrm{Cl^-}) m(\mathrm{sample})^{-1} \times 100~\%\\[1.5ex] &= \dfrac{0.2010~\mathrm{mol~AgNO_3}}{\mathrm{L}} \left ( \dfrac{70.90~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \left ( \dfrac{\mathrm{mol~Ag^+}}{\mathrm{mol~AgNO_3}} \right ) \left ( \dfrac{\mathrm{mol~AgCl}}{\mathrm{mol~Ag^+}} \right ) \left ( \dfrac{\mathrm{mol~Cl^-}}{\mathrm{mol~AgCl}} \right )\\ &\phantom{=}~~ \left ( \dfrac{35.45~\mathrm{g}}{\mathrm{mol~Cl^-}} \right ) \left ( \dfrac{1}{4.106~\mathrm{g}} \right ) \times 100~\% \\[1.5ex] &= 12.3\bar{0}38~\% = 12.30~\% \end{align*}\]
Concept: stoichiometry; molarity; percent composition by mass
A mixture of BaCl2 and NaCl is analyzed by precipitating all of the barium as BaSO4. After the addition of excess Na2SO4 to a 3.988 g sample of the mixture, the mass of precipitate collected is 2.113 g. What is the mass percentage of barium chloride in the mixture?
Solution
Answer: 47.27 %
\[\begin{align*} w(\mathrm{BaCl_2})\% &= w(\mathrm{BaCl_2}) \times 100~\%\\[1.5ex] &= m(\mathrm{BaCl_2}) ~ m(\mathrm{sample})^{-1} \times 100~\%\\[1.5ex] &= n(\mathrm{BaCl_2}) ~ M(\mathrm{BaCl_2}) ~ m(\mathrm{sample})^{-1} \times 100~\%\\[1.5ex] &= n(\mathrm{BaSO_4}) ~ r(\mathrm{BaCl_2,BaSO_4}) ~ M(\mathrm{BaCl_2}) ~ m(\mathrm{sample})^{-1} \times 100~\%\\[1.5ex] &= m(\mathrm{BaSO_4}) ~ M(\mathrm{BaSO_4})^{-1} ~ r(\mathrm{BaCl_2,BaSO_4}) ~ M(\mathrm{BaCl_2}) ~ m(\mathrm{sample})^{-1} \times 100~\%\\[1.5ex] &= 2.113~\mathrm{g~BaSO_4} \left ( \dfrac{\mathrm{mol~BaSO_4}}{233.39~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{mol~BaCl_2}}{\mathrm{mol~BaSO_4}} \right ) \left ( \dfrac{208.23~\mathrm{g}}{\mathrm{mol~BaCl_2}} \right ) \left ( \dfrac{1}{3.988~\mathrm{g}} \right ) \times 100~\% \\[1.5ex] &= 47.2\bar{7}21~\% = 47.27~\% \end{align*}\]
Concept: stoichiometry; molarity; percent composition by mass
A 3.00 g sample of an alloy containing only Pb and Sn was dissolved in nitric acid. Sulfuric acid was added to this solution, which precipitated 1.90 g of PbSO4. Assuming that all of the lead was precipitated, what is the percentage of Sn in the sample? (M(PbSO4) = 303.26 g mol–1)
Solution
Answer: 56.7 %
\[\begin{align*} \mathrm{alloy} + \mathrm{HNO_3(aq)} + \mathrm{H_2SO_4(aq)} \longrightarrow \mathrm{PbSO_4(s)} + \mathrm{other~stuff} \end{align*}\]
\[\begin{align*} w(\mathrm{Sn})\% &= w(\mathrm{Sn}) \times 100~\%\\[1.5ex] &= m(\mathrm{Sn}) ~ m(\mathrm{sample})^{-1} \times 100~\%\\[1.5ex] &= \biggl\{ m(\mathrm{sample}) - m(\mathrm{Pb^{2+}}) \biggl\}~ m(\mathrm{sample})^{-1} \times 100~\% \\[1.5ex] &= \biggl\{ m(\mathrm{sample}) - \biggr[ n(\mathrm{Pb^{2+}})~ M(\mathrm{Pb^{2+}}) \biggr] \biggl\}~ m(\mathrm{sample})^{-1} \times 100~\% \\[1.5ex] &= \Biggl\{ m(\mathrm{sample}) - \biggr[ n(\mathrm{PbSO_4}) ~ r(\mathrm{Pb^{2+},PbSO_4}) ~ M(\mathrm{Pb^{2+}}) \biggr] \Biggl\} ~ m(\mathrm{sample})^{-1} \times 100~\%\\[1.5ex] &= \Biggl\{ m(\mathrm{sample}) - \biggr[ m(\mathrm{PbSO_4}) ~ M(\mathrm{PbSO_4})^{-1} ~ r(\mathrm{Pb^{2+},PbSO_4}) ~ M(\mathrm{Pb^{2+}}) \biggr] \Biggl\} ~ \\ &\phantom{=}~~~~~ m(\mathrm{sample})^{-1} \times 100~\%\\[1.5ex] &= \Biggl\{ 3.00~\mathrm{g~sample} - \left [ 1.90~\mathrm{g~PbSO_4} \left ( \dfrac{\mathrm{mol~PbSO_4}}{303.26~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{mol~Pb^{2+}}}{\mathrm{mol~PbSO_4}} \right ) \left ( \dfrac{207.2~\mathrm{g}}{\mathrm{mol~Pb^{2+}}} \right ) \right ] \Biggl\} \\ &\phantom{=}~~ \left ( \dfrac{1}{3.00~\mathrm{g~sample}} \right ) \times 100~\% \\[1.5ex] &= 56.\bar{7}28~\% = 56.7~\% \end{align*}\]
Concept: stoichiometry; molarity; percent composition by mass
You have 76.0 mL of a 2.50 M aqueous solution of Na2CrO4 and 125 mL of a 2.16 M aqueous solution of AgNO3. Calculate the molar concentration (in mol L–1) of CrO42– after the two solutions are mixed together.
Solution
Answer: 0.274 M
\[\begin{align*} \mathrm{Na_2CrO_4(aq)} + \mathrm{2~AgNO_3(aq)} \longrightarrow \mathrm{Ag_2CrO_4(s)} + \mathrm{2~NaNO_3(aq)} \end{align*}\]
Find the limiting reactant.
\[\begin{align*} n(\mathrm{Ag_2CrO_4}) &= n(\mathrm{Na_2CrO_4}) ~ r(\mathrm{Ag_2CrO_4,Na_2CrO_4}) \\[1.5ex] &= c(\mathrm{Na_2CrO_4}) ~ V(\mathrm{Na_2CrO_4}) ~ r(\mathrm{Ag_2CrO_4,Na_2CrO_4}) \\[1.5ex] &= \dfrac{2.50~\mathrm{mol~Na_2CrO_4}}{\mathrm{L}} \left ( \dfrac{76.0~\mathrm{mL~Na_2CrO_4}}{} \right ) \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \left ( \dfrac{\mathrm{mol~Ag_2CrO_4}}{\mathrm{mol~Na_2CrO_4}} \right )\\[1.5ex] &= 0.19\bar{0}00~\mathrm{mol} \\[3.0ex] n(\mathrm{Ag_2CrO_4}) &= n(\mathrm{AgNO_3}) ~ r(\mathrm{Ag_2CrO_4,AgNO_3}) \\[1.5ex] &= c(\mathrm{AgNO_3}) ~ V(\mathrm{AgNO_3}) ~ r(\mathrm{Ag_2CrO_4,AgNO_3}) \\[1.5ex] &= \dfrac{2.16~\mathrm{mol~AgNO_3}}{\mathrm{L}} \left ( \dfrac{125~\mathrm{mL~AgNO_3}}{} \right ) \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \left ( \dfrac{\mathrm{mol~Ag_2CrO_4}}{\mathrm{2~mol~AgNO_3}} \right ) \\[1.5ex] &= 0.13\bar{5}00~\mathrm{mol} \end{align*}\]
AgNO3 is limiting.
\[\begin{align*} c(\mathrm{CrO_4^{2-}}) &= n(\mathrm{CrO_4^{2-}}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= n(\mathrm{Na_2CrO_4})_{\mathrm{leftover}} ~ r(\mathrm{CrO_4^{2-},Na_2CrO_4}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ n(\mathrm{Na_2CrO_4})_{\mathrm{initial}} - n(\mathrm{Na_2CrO_4})_{\mathrm{reacted}} \biggl\} ~ r(\mathrm{CrO_4^{2-},Na_2CrO_4}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \Biggl\{ \biggr[ V(\mathrm{Na_2CrO_4}) ~ c(\mathrm{Na_2CrO_4}) \biggr] ~ - ~ \biggr[ n(\mathrm{Ag_2CrO_4})_{\mathrm{formed}} ~ r(\mathrm{Na_2CrO_4,Ag_2CrO_4}) ~ \biggr] \Biggl\} \\ &\phantom{=}~~~~ r(\mathrm{CrO_4^{2-},AgNO_3}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ \biggr[ 76.0~\mathrm{mL~Na_2CrO_4} \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \left ( \dfrac{2.50~\mathrm{mol~Na_2CrO_4}}{\mathrm{L}} \right ) \biggr] ~-~ \\ &\phantom{=\biggl\{~} \biggr[ 0.13\bar{5}00~\mathrm{mol~AgCrO_4} \left ( \dfrac{\mathrm{mol~Na_2CrO_4}}{\mathrm{mol~Ag_2CrO_4}} \right ) \biggr] \biggl\} \\ &\phantom{=(} \left ( \dfrac{\mathrm{mol~CrO_4^{2-}}}{\mathrm{mol~Na_2CrO_4}} \right ) \left ( \dfrac{1}{(76.0 + 125)~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 0.27\bar{3}63~\mathrm{mol~L^{-1}} \\[1.5ex] &= 0.274~\mathrm{mol~L^{-1}} \end{align*}\]
Concept: balancing equations; stoichiometry; molarity; solubility rules; limiting reactant
You have a 75.0 mL 2.50 M aqueous Na2CrO4 solution and 125 mL 2.29 M aqueous AgNO3 solution. What is the molar concentration (in mol L–1) of Ag+ after the two solutions are mixed together?
Solution
Answer: 0 M
\[\begin{align*} \mathrm{Na_2CrO_4(aq)} + \mathrm{2~AgNO_3(aq)} \longrightarrow \mathrm{Ag_2CrO_4(s)} + \mathrm{2~NaNO_3(aq)} \end{align*}\]
Find the limiting reactant.
\[\begin{align*} n(\mathrm{Ag_2CrO_4}) &= n(\mathrm{Na_2CrO_4}) ~ r(\mathrm{Ag_2CrO_4,Na_2CrO_4}) \\[1.5ex] &= V(\mathrm{Na_2CrO_4}) ~ c(\mathrm{Na_2CrO_4}) ~ r(\mathrm{Ag_2CrO_4,Na_2CrO_4}) \\[1.5ex] &= 75.0~\mathrm{mL~Na_2CrO_4} \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \left ( \dfrac{2.50~\mathrm{mol~Na_2CrO_4}}{\mathrm{L}} \right ) \left ( \dfrac{\mathrm{mol~Ag_2CrO_4}}{\mathrm{mol~Na_2CrO_4}} \right )\\[1.5ex] &= 0.18\bar{7}50~\mathrm{mol} \\[3ex] n(\mathrm{Ag_2CrO_4}) &= n(\mathrm{AgNO_3}) ~ r(\mathrm{Ag_2CrO_4,AgNO_3}) \\[1.5ex] &= V(\mathrm{AgNO_3}) ~ c(\mathrm{AgNO_3}) ~ r(\mathrm{Ag_2CrO_4,AgNO_3}) \\[1.5ex] &= 125~\mathrm{mL~AgNO_3} \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \left ( \dfrac{2.24~\mathrm{mol~AgNO_3}}{\mathrm{L}} \right ) \left ( \dfrac{\mathrm{mol~Ag_2CrO_4}}{\mathrm{2~mol~AgNO_3}} \right ) \\[1.5ex] &= 0.14\bar{0}00~\mathrm{mol} \end{align*}\]
AgNO3 is limiting.
\[\begin{align*} c(\mathrm{NO_3^{2-}}) &= n(\mathrm{NO_3^{2-}}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= n(\mathrm{AgNO_3}) ~ r(\mathrm{NO_3^{2-},AgNO_3}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= n(\mathrm{Ag_2CrO_4}) ~ r(\mathrm{AgNO_3,Ag_2CrO_4}) ~ r(\mathrm{NO_3^{2-},AgNO_3}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= 0.14\bar{0}00~\mathrm{mol} \left ( \dfrac{\mathrm{2~mol~AgNO_3}}{\mathrm{mol~Ag_2CrO_4}} \right ) \left ( \dfrac{\mathrm{mol~NO_3^{2-}}}{\mathrm{mol~AgNO_3}} \right ) \left ( \dfrac{1}{(75.0+125)~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 1.4\bar{0}00~\mathrm{mol~L^{-1}} = 1.40~\mathrm{mol~L^{-1}} \end{align*}\]
AgNO3 is limiting; therefore, all Ag+ is converted to Ag2CrO4(s).
\[\begin{align*} c(\mathrm{Ag}^+) &= 0~\mathrm{mol~L^{-1}} \end{align*}\]
Concept: balancing equations; stoichiometry; molarity; solubility rules; limiting reactant
You have 75.0 mL of a 2.50 M aqueous solution of Na2CrO4 and 125 mL of a 2.24 M aqueous solution of AgNO3. Calculate the molar concentration (in mol L–1) of NO3– after the two solutions are mixed together.
Solution
Answer: 0 M
\[\begin{align*} \mathrm{Na_2CrO_4(aq)} + \mathrm{2~AgNO_3(aq)} \longrightarrow \mathrm{Ag_2CrO_4(s)} + \mathrm{2~NaNO_3(aq)} \end{align*}\]
Find the limiting reactant.
\[\begin{align*} n(\mathrm{Ag_2CrO_4}) &= n(\mathrm{Na_2CrO_4}) ~ r(\mathrm{Ag_2CrO_4,Na_2CrO_4}) \\[1.5ex] &= V(\mathrm{Na_2CrO_4}) ~ c(\mathrm{Na_2CrO_4}) ~ r(\mathrm{Ag_2CrO_4,Na_2CrO_4}) \\[1.5ex] &= 75.0~\mathrm{mL~Na_2CrO_4} \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \left ( \dfrac{2.50~\mathrm{mol~Na_2CrO_4}}{\mathrm{L}} \right ) \left ( \dfrac{\mathrm{mol~Ag_2CrO_4}}{\mathrm{mol~Na_2CrO_4}} \right )\\[1.5ex] &= 0.18\bar{7}50~\mathrm{mol} \\[3ex] n(\mathrm{Ag_2CrO_4}) &= n(\mathrm{AgNO_3}) ~ r(\mathrm{Ag_2CrO_4,AgNO_3}) \\[1.5ex] &= V(\mathrm{AgNO_3}) ~ c(\mathrm{AgNO_3}) ~ r(\mathrm{Ag_2CrO_4,AgNO_3}) \\[1.5ex] &= 125~\mathrm{mL~AgNO_3} \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \left ( \dfrac{2.29~\mathrm{mol~AgNO_3}}{\mathrm{L}} \right ) \left ( \dfrac{\mathrm{mol~Ag_2CrO_4}}{\mathrm{2~mol~AgNO_3}} \right ) \\[1.5ex] &= 0.14\bar{3}12~\mathrm{mol} \end{align*}\]
AgNO3 is limiting.
\[\begin{align*} c(\mathrm{CrO_4^{2-}}) &= n(\mathrm{CrO_4^{2-}}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= n(\mathrm{Na_2CrO_4})_{\mathrm{leftover}} ~ r(\mathrm{CrO_4^{2-},Na_2CrO_4}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ n(\mathrm{Na_2CrO_4})_{\mathrm{initial}} - n(\mathrm{Na_2CrO_4})_{\mathrm{reacted}} \biggl\} ~ r(\mathrm{CrO_4^{2-},Na_2CrO_4}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \Biggl\{ \biggr[ V(\mathrm{Na_2CrO_4}) ~ c(\mathrm{Na_2CrO_4}) \biggr] ~ - ~ \biggr[ n(\mathrm{Ag_2CrO_4})_{\mathrm{formed}} ~ r(\mathrm{Na_2CrO_4,Ag_2CrO_4}) ~ \biggr] \Biggl\} \\ &\phantom{=}~~~~ r(\mathrm{CrO_4^{2-},AgNO_3}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ \biggr[ 76.0~\mathrm{mL~Na_2CrO_4} \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \left ( \dfrac{2.50~\mathrm{mol~Na_2CrO_4}}{\mathrm{L}} \right ) \biggr] ~-~ \\ &\phantom{=\biggl\{~} \biggr[ 0.13\bar{5}00~\mathrm{mol~AgCrO_4} \left ( \dfrac{\mathrm{mol~Na_2CrO_4}}{\mathrm{mol~Ag_2CrO_4}} \right ) \biggr] \biggl\} \\ &\phantom{=(} \left ( \dfrac{\mathrm{mol~CrO_4^{2-}}}{\mathrm{mol~Na_2CrO_4}} \right ) \left ( \dfrac{1}{(76.0 + 125)~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 0.27\bar{3}63~\mathrm{mol~L^{-1}} \\[1.5ex] &= 0.274~\mathrm{mol~L^{-1}} \end{align*}\]
AgNO3 is limiting.
\[\begin{align*} c(\mathrm{Ag}^+) &= 0~\mathrm{mol~L^{-1}} \end{align*}\]
Concept: balancing equations; stoichiometry; molarity; solubility rules; limiting reactant
Combine a 55 mL 1.00 M aqueous silver nitrate solution with a 25 mL 0.55 M silver chloride solution. What mass (in g) of silver chloride is produced?
Solution
Answer: 2.0 g
\[\begin{align*} \mathrm{AgNO_3(aq)} + \mathrm{NaCl(aq)} \longrightarrow \mathrm{AgCl(s)} + \mathrm{NaNO_3(aq)} \end{align*}\]
Find the limiting reactant.
\[\begin{align*} n(\mathrm{AgCl}) &= n(\mathrm{AgNO_3}) ~ r(\mathrm{AgCl,AgNO_3}) \\[1.5ex] &= V(\mathrm{AgNO_3}) ~ c(\mathrm{AgNO_3}) ~ r(\mathrm{AgCl,AgNO_3}) \\[1.5ex] &= 55~\mathrm{mL~AgNO_3} \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \left ( \dfrac{1.00~\mathrm{mol~AgNO_3}}{\mathrm{L}} \right ) \left ( \dfrac{\mathrm{mol~AgCl}}{\mathrm{mol~AgNO_3}} \right )\\[1.5ex] &= 0.05\bar{5}00~\mathrm{mol} \\[3ex] n(\mathrm{AgCl}) &= n(\mathrm{NaCl}) ~ r(\mathrm{AgCl,NaCl}) \\[1.5ex] &= V(\mathrm{NaCl}) ~ c(\mathrm{NaCl}) ~ r(\mathrm{AgCl,NaCl}) \\[1.5ex] &= 25~\mathrm{mL~NaCl} \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \left ( \dfrac{0.55~\mathrm{mol~NaCl}}{\mathrm{L}} \right ) \left ( \dfrac{\mathrm{mol~AgCl}}{\mathrm{mol~NaCl}} \right )\\[1.5ex] &= 0.01\bar{3}75~\mathrm{mol} \end{align*}\]
NaCl is limiting.
\[\begin{align*} m(\mathrm{AgCl}) &= n(\mathrm{AgCl}) ~ M(\mathrm{AgCl})^{-1} \\[1.5ex] &= 0.01\bar{3}75~\mathrm{mol} \left ( \dfrac{143.32~\mathrm{g}}{\mathrm{mol~AgCl}} \right ) \\[1.5ex] &= 1.\bar{9}70~\mathrm{g} = 2.0~\mathrm{g} \end{align*}\]
Concept: balancing equations; stoichiometry; molarity; solubility rules; limiting reactant
A 0.685 g sample of an unknown diprotic acid requires a 42.57 mL 0.111 M aqueous NaOH solution to be completely neutralized. What is the molar mass (in g mol–1) of the acid?
Solution
Answer: 290. g mol–1
\[\begin{align*} \mathrm{H_2X(aq)} + \mathrm{2~NaOH(aq)} \longrightarrow \mathrm{2~H_2O(l)} + \mathrm{Na_2X(aq)} \end{align*}\]
\[\begin{align*} M(\mathrm{H_2X}) &= m(\mathrm{H_2X}) ~ n(\mathrm{H_2X})^{-1} \\[1.5ex] &= m(\mathrm{H_2X}) ~ \biggr[ n(\mathrm{NaOH}) ~ r(\mathrm{H_2X,NaOH}) \biggr]^{-1} \\[1.5ex] &= m(\mathrm{H_2X}) ~ \biggr[ c(\mathrm{NaOH}) ~ V(\mathrm{solution}) ~ r(\mathrm{H_2X,NaOH}) \biggr]^{-1} \\[1.5ex] &= 0.685~\mathrm{g~H_2X} ~ \biggr[ \left ( \dfrac{0.111~\mathrm{mol~NaOH}}{\mathrm{L}} \right ) \left ( \dfrac{42.57~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \left ( \dfrac{\mathrm{mol~H_2X}}{2~\mathrm{mol~NaOH}} \right ) \biggr]^{-1} \\[1.5ex] &= 28\bar{9}.93~\mathrm{g~mol^{-1}} = 290.~\mathrm{g~mol^{-1}} \end{align*}\]
Concept: balancing equations; stoichiometry; molarity; acid-base reaction
What mass (in g) of NaOH is required to completely react with a 25.0 mL 2.2 M aqueous H2SO4 solution?
Solution
Answer: 290. g mol–1
\[\begin{align*} \mathrm{H_2SO_4(aq)} + \mathrm{2~NaOH(aq)} \longrightarrow \mathrm{2~H_2O(l)} + \mathrm{Na_2SO_4(aq)} \end{align*}\]
\[\begin{align*} m(\mathrm{NaOH}) &= n(\mathrm{NaOH}) ~ M(\mathrm{NaOH}) \\[1.5ex] &= n(\mathrm{H_2SO_4}) ~ r(\mathrm{NaOH,H_2SO_4}) ~ M(\mathrm{NaOH}) \\[1.5ex] &= c(\mathrm{H_2SO_4}) ~ V(\mathrm{H_2SO_4}) ~ r(\mathrm{NaOH,H_2SO_4}) ~ M(\mathrm{NaOH}) \\[1.5ex] &= \dfrac{2.2~\mathrm{mol~H_2SO_4}}{\mathrm{L}} \left ( \dfrac{25.0~\mathrm{mL~H_2SO_4}}{} \right ) \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \left ( \dfrac{2~\mathrm{mol~NaOH}}{\mathrm{mol~H_2SO_4}} \right ) \left ( \dfrac{40.00~\mathrm{g}}{\mathrm{mol~NaOH}} \right ) \\[1.5ex] &= 4.\bar{4}00~\mathrm{g} = 4.4~\mathrm{g} \end{align*}\]
Concept: balancing equations; stoichiometry; molarity; acid-base reaction
What volume (in mL) of a 5.00 M hydrofluoric acid solution will completely react with 4.05 g of calcium hydroxide?
Solution
Answer: 21.9 mL
\[\begin{align*} \mathrm{HF(aq)} + \mathrm{Ca(OH)_2} \longrightarrow \mathrm{2~H_2O(l)} + \mathrm{CaF_2(aq)} \end{align*}\]
\[\begin{align*} V(\mathrm{HF}) &= n(\mathrm{HF}) ~ c(\mathrm{HF})^{-1} \\[1.5ex] &= n(\mathrm{Ca(OH)_2}) ~ r(\mathrm{HF,Ca(OH)_2}) ~ c(\mathrm{HF})^{-1} \\[1.5ex] &= m(\mathrm{Ca(OH)_2}) ~ M(\mathrm{Ca(OH)_2})^{-1} ~ r(\mathrm{HF,Ca(OH)_2}) ~ c(\mathrm{HF})^{-1} \\[1.5ex] &= 4.05~\mathrm{g~Ca(OH)_2} \left ( \dfrac{\mathrm{mol~Ca(OH)_2}}{74.10~\mathrm{g}} \right ) \left ( \dfrac{2~\mathrm{mol~HF}}{\mathrm{mol~Ca(OH)_2}} \right ) \left ( \dfrac{\mathrm{L}}{\mathrm{5.00~mol~HF}} \right ) \left ( \dfrac{\mathrm{10^3~mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 21.\bar{8}62~\mathrm{mL} = 21.9~\mathrm{mL} \end{align*}\]
Concept: balancing equations; stoichiometry; molarity; acid-base reaction
Sulfamic acid (HSO3NH2) is a strong monoprotic acid that can be used to standardize a strong base. A 0.179 g sample of HSO3NH2 is required to completely neutralize a 19.4 mL aqueous KOH solution. What is the molar concentration (in mol L–1) of the KOH solution?
Solution
Answer: 0.0950 M
\[\begin{align*} \mathrm{HSO_3NH_3(aq)} + \mathrm{KOH} \longrightarrow \mathrm{H_2O(l)} + \mathrm{KSO_3NH_2(aq)} \end{align*}\]
\[\begin{align*} c(\mathrm{KOH}) &= n(\mathrm{KOH}) ~ V(\mathrm{KOH})^{-1} \\[1.5ex] &= n(\mathrm{HSO_3NH_2}) ~ r(\mathrm{KOH,HSO_3NH_2}) ~ V(\mathrm{KOH})^{-1} \\[1.5ex] &= m(\mathrm{HSO_3NH_2}) ~ M(\mathrm{HSO_3NH_2})^{-1} ~ r(\mathrm{KOH,HSO_3NH_2}) ~ V(\mathrm{KOH})^{-1} \\[1.5ex] &= 0.179~\mathrm{g~HSO_3NH_2} \left ( \dfrac{\mathrm{mol~HSO_3NH_2}}{97.10~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{mol~KOH}}{\mathrm{mol~HSO_3NH_2}} \right ) \left ( \dfrac{1}{\mathrm{19.4~mL~KOH}} \right ) \left ( \dfrac{\mathrm{10^3~mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 0.095\bar{0}23~\mathrm{mol~L^{-1}} = 0.0950~\mathrm{mol~L^{-1}} \end{align*}\]
Concept: balancing equations; stoichiometry; molarity; acid-base reaction
A student weighs out 0.556 g of KHP (M(KHP) = 204.22 g mol–1) and puts it into 36.78 mL of a stock NaOH solution. If that was enough to neutralize the NaOH, what is the concentration of the stock NaOH solution? KHP is a monoprotic acid.
Solution
Answer: 0.0740 M
\[\begin{align*} \mathrm{HX(aq)} + \mathrm{NaOH} \longrightarrow \mathrm{H_2O(l)} + \mathrm{KSO_3NH_2(aq)} \end{align*}\]
\[\begin{align*} c(\mathrm{NaOH}) &= n(\mathrm{NaOH}) ~ V(\mathrm{NaOH})^{-1} \\[1.5ex] &= n(\mathrm{HX}) ~ r(\mathrm{NaOH,HX}) ~ V(\mathrm{NaOH})^{-1} \\[1.5ex] &= m(\mathrm{HX}) ~ M(\mathrm{HX})^{-1} ~ r(\mathrm{NaOH,HX}) ~ V(\mathrm{NaOH})^{-1} \\[1.5ex] &= 0.556~\mathrm{g~HX} \left ( \dfrac{\mathrm{mol~HX}}{204.22~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{mol~NaOH}}{\mathrm{mol~HX}} \right ) \left ( \dfrac{1}{\mathrm{36.78~mL~NaOH}} \right ) \left ( \dfrac{\mathrm{10^3~mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 0.074\bar{0}22~\mathrm{mol~L^{-1}} = 0.0740~\mathrm{mol~L^{-1}} \end{align*}\]
Concept: balancing equations; stoichiometry; molarity; acid-base reaction
A 2.80 g sample of phosphoric acid is added to a 150.0 mL 1.00 M sodium hydroxide solution to give a 151.489 mL mixture and the acid is completely neutralized. Determine the following:
- [Na+] (in mol L–1)
- [PO43–] (in mol L–1)
- [OH–] (in mol L–1)
Solution
Answer: 1.00 M; 0.189 M; 0.424 M
\[\begin{align*} \mathrm{H_3PO_4(aq)} + \mathrm{3~NaOH} \longrightarrow \mathrm{3~H_2O(l)} + \mathrm{Na_3PO_4(aq)}\\ \end{align*}\]
Determine the limiting reagent.
\[\begin{align*} n(\mathrm{Na_3PO_4}) &= n(\mathrm{H_3PO_4}) ~ r(\mathrm{Na_3PO_4,H_3PO_4}) \\[1.5ex] &= m(\mathrm{H_3PO_4}) ~ M(\mathrm{H_3PO_4})^{-1} ~ r(\mathrm{Na_3PO_4}) \\[1.5ex] &= 2.80~\mathrm{g~H_3PO_4} \left ( \dfrac{\mathrm{mol~H_3PO_4}}{\mathrm{98.00~g}} \right ) \left ( \dfrac{\mathrm{mol~Na_3PO_4}}{\mathrm{mol~H_3PO_4}} \right )\\[1.5ex] &= 0.028\bar{5}71~\mathrm{mol} \\[3ex] n(\mathrm{Na_3PO_4}) &= n(\mathrm{NaOH}) ~ r(\mathrm{Na_3PO_4,NaOH}) \\[1.5ex] &= c(\mathrm{NaOH}) ~ V(\mathrm{NaOH}) ~ r(\mathrm{Na_3PO_4,NaOH}) \\[1.5ex] &= \dfrac{1.00~\mathrm{mol~NaOH}}{\mathrm{L}} \left ( \dfrac{150.0~\mathrm{mL~NaOH}}{} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \left ( \dfrac{\mathrm{mol~Na_3PO_4}}{\mathrm{3~mol~NaOH}} \right ) \\[1.5ex] &= 0.050\bar{0}00~\mathrm{mol} \end{align*}\]
H3PO4 is limiting.
A
Since NaOH and Na3PO3 are strong electrolytes, the Na+ concentration is equal to the initial NaOH(aq) concentration. [Na+] = 1.00 mol L–1
B
\[\begin{align*} c(\mathrm{PO_4^{3-}}) &= n(\mathrm{PO_4^{3-}}) ~ V(\mathrm{solution})^{-1}\\[1.5ex] &= n(\mathrm{Na_3PO_4}) ~ r(\mathrm{PO_4^{3-},Na_3PO_4}) ~ V(\mathrm{solution}^{-1})\\[1.5ex] &= 0.028\bar{5}71~\mathrm{mol~Na_3PO_4} \left ( \dfrac{\mathrm{mol~PO_4^{3-}}}{\mathrm{mol~Na_3PO_4}} \right ) \left ( \dfrac{1}{\mathrm{151.489~\mathrm{mL}}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 0.18\bar{8}60~\mathrm{mol~L^{-1}} = 0.189~\mathrm{mol~L^{-1}} \end{align*}\]
C
\[\begin{align*} c(\mathrm{OH^-}) &= n(\mathrm{OH^-}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ n(\mathrm{NaOH})_{\mathrm{initial}} - n(\mathrm{NaOH})_{\mathrm{consumed}} \biggl\} ~ r(\mathrm{OH^-,NaOH}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ \biggr[ c(\mathrm{NaOH})~ V(\mathrm{NaOH}) \biggr] ~ - ~ \biggr[ n(\mathrm{Na_3PO_4}) ~ r(\mathrm{NaOH,Na_3PO_4}) \biggr] \biggl\} ~ r(\mathrm{OH^-,NaOH}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ \biggr[ \dfrac{1.00~\mathrm{mol~NaOH}}{\mathrm{L}} \left ( \dfrac{150.0~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \biggr] ~-\\ &\phantom{=}~~~ \biggr[ 0.028\bar{5}71~\mathrm{mol~Na_3PO_4} \left ( \dfrac{3~\mathrm{mol~NaOH}}{\mathrm{Na_3PO_4}} \right ) \biggr] \biggl\} \\ &\phantom{=}~~~ \left ( \dfrac{\mathrm{mol~OH^-}}{\mathrm{mol~NaOH}} \right ) \left ( \dfrac{1}{\mathrm{151.489~mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 0.42\bar{4}36~\mathrm{mol~L^{-1}} = 0.424~\mathrm{mol~L^{-1}} \end{align*}\]
Concept: balancing equations; stoichiometry; molarity; acid-base reaction; limiting reactant
A stock solution with a total volume of 1000.0 mL contains 37.1 g Mg(NO3)2. If you take a 20.0 mL aliquot and then dilute it with water to a total volume of 500.0 mL, what is the molar concentration (in g mol–1) of Mg2+ and NO32– in the final solution?
Solution
Answer: 0.0100 M; 0.0200 M
\[\begin{align*} c(\mathrm{Mg^{2+}}) &= n(\mathrm{Mg^{2+}})_{\textrm{dilute aliquot}} ~ V(\mathrm{dilute~aliquot})^{-1} \\[1.5ex] &= n(\mathrm{Mg^{2+}})_{\textrm{stock solution}} ~ V(\mathrm{stock~solution})^{-1} ~ V(\mathrm{aliquot}) ~ V(\mathrm{dilute~aliquot})^{-1} \\[1.5ex] &= m(\mathrm{Mg(NO_3)_2}) ~ M(\mathrm{Mg(NO_3)_2})^{-1} ~ r(\mathrm{Mg^{2+},Mg(NO_3)_2})\\ &\phantom{=}~~ V(\mathrm{stock~solution})^{-1} ~ V(\mathrm{aliquot}) ~ V(\mathrm{dilute~aliquot})^{-1} \\[1.5ex] &= 37.1~\mathrm{g~Mg(NO_3)_2} \left ( \dfrac{\mathrm{mol~Mg(NO_3)_2}}{148.33~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{mol~Mg^{2+}}}{\mathrm{mol~Mg(NO_3)_2}} \right ) \\ &\phantom{=(} \left ( \dfrac{1}{1000.0~\mathrm{mL}} \right ) \left ( \dfrac{20.0~\mathrm{mL}}{} \right ) \left ( \dfrac{1}{500.0~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 0.010\bar{0}04~\mathrm{mol~L^{-1}} = 0.0100~\mathrm{mol~L^{-1}} \\[3ex] c(\mathrm{NO_3^{2-}}) &= n(\mathrm{NO_3^{2-}})_{\textrm{dilute~aliquot}} ~ V(\mathrm{stock~solution})^{-1} \\[1.5ex] &= n(\mathrm{NO_3^{2-}})_{\textrm{stock~solution}} ~ V(\mathrm{stock~solution})^{-1}~ V(\mathrm{aliquot}) ~ V(\mathrm{dilute~aliquot})^{-1} \\[1.5ex] &= m(\mathrm{Mg(NO_3)_2}) ~ M(\mathrm{Mg(NO_3)_2})^{-1} ~ r(\mathrm{NO_3^{2-},Mg(NO_3)_2})\\ &\phantom{=}~~ V(\mathrm{stock~solution})^{-1} ~ V(\mathrm{aliquot}) ~ V(\mathrm{dilute~aliquot})^{-1} \\[1.5ex] &= 37.1~\mathrm{g~Mg(NO_3)_2} \left ( \dfrac{\mathrm{mol~Mg(NO_3)_2}}{148.33~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{2~mol~NO_3^{2-}}}{\mathrm{mol~Mg(NO_3)_2}} \right ) \\ &\phantom{=(} \left ( \dfrac{1}{1000.0~\mathrm{mL}} \right ) \left ( \dfrac{20.0~\mathrm{mL}}{} \right ) \left ( \dfrac{1}{500.0~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 0.010\bar{0}04~\mathrm{mol~L^{-1}} = 0.0100~\mathrm{mol~L^{-1}} \\[3ex] c(\mathrm{NO_3^{2-}}) &= n(\mathrm{NO_3^{2-}})_{\textrm{dilute~aliquot}} ~ V(\mathrm{stock~solution})^{-1} \\[1.5ex] &= n(\mathrm{NO_3^{2-}})_{\textrm{stock~solution}} ~ V(\mathrm{stock~solution})^{-1}~ V(\mathrm{aliquot}) ~ V(\mathrm{dilute~aliquot})^{-1} \\[1.5ex] &= m(\mathrm{Mg(NO_3)_2}) ~ M(\mathrm{Mg(NO_3)_2})^{-1} ~ r(\mathrm{NO_3^{2-},Mg(NO_3)_2})\\ &\phantom{=}~~ V(\mathrm{stock~solution})^{-1} ~ V(\mathrm{aliquot}) ~ V(\mathrm{dilute~aliquot})^{-1} \\[1.5ex] &= 37.1~\mathrm{g~Mg(NO_3)_2} \left ( \dfrac{\mathrm{mol~Mg(NO_3)_2}}{148.33~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{2~mol~NO_3^{2-}}}{\mathrm{mol~Mg(NO_3)_2}} \right ) \\ &\phantom{=(} \left ( \dfrac{1}{1000.0~\mathrm{mL}} \right ) \left ( \dfrac{20.0~\mathrm{mL}}{} \right ) \left ( \dfrac{1}{500.0~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 0.020\bar{0}09~\mathrm{mol~L^{-1}} = 0.0200~\mathrm{mol~L^{-1}} \end{align*}\]
Alternatively,
\[\begin{align*} c(\mathrm{NO_3^{2-}}) &= c(\mathrm{Mg^{2+}})_{\textrm{dilute~aliquot}} ~ r(\mathrm{NO_3^{2-},Mg^{2+}}) \\[1.5ex] &= \dfrac{0.010\bar{0}04~\mathrm{mol}}{\mathrm{L}} \left ( \dfrac{2~\mathrm{mol~NO_3^-}}{\mathrm{mol~Mg(NO_3)_2}} \right ) \\[1.5ex] &= 0.020\bar{0}08~\mathrm{mol~L^{-1}} = 0.0200~\mathrm{mol~L^{-1}} \end{align*}\]
Concept: stoichiometry; molarity; dilution
Determine the molar concentrations (in g mol–1) the ions present in a solution created from mixing equal volumes of 1.0 M aqueous lead(II) nitrate and 1.0 M aqueous sodium chloride solutions? Assume that the volumes are precise to one decimal place in normalized scientific notation.
Solution
Answer: Pb2+: 0.25 M; NO32–: 1.0 M; Na+: 0.50 M; Cl–: 0 M
This solution considers the mixing of two 1.0 × 100 L solutions to give a 2.0 L final solution.
Determine the limiting reactant.
\[\begin{align*} n(\mathrm{PbCl_2}) &= c(\mathrm{Pb(NO_3)_2}) ~ V(\mathrm{Pb(NO_3)_2})_{\mathrm{initial}} ~ r(\mathrm{PbCl_2,Pb(NO_3)_2}) \\[1.5ex] &= \dfrac{1.0~\mathrm{mol~Pb(NO_3)_2}}{\mathrm{L}} \left ( \dfrac{1.0~\mathrm{L}}{} \right ) \left ( \dfrac{\mathrm{mol~PbCl_2}}{\mathrm{mol~Pb(NO_3)_2}} \right ) \\[1.5ex] &= 1.0~\mathrm{mol} \\[3ex] n(\mathrm{PbCl_2}) &= c(\mathrm{NaCl}) ~ V(\mathrm{NaCl})_{\mathrm{initial}} ~ r(\mathrm{PbCl_2,NaCl}) \\[1.5ex] &= \dfrac{1.0~\mathrm{mol~NaCl}}{\mathrm{L}} \left ( \dfrac{1.0~\mathrm{L}}{} \right ) \left ( \dfrac{\mathrm{mol~PbCl_2}}{2~\mathrm{mol~NaCl}} \right ) \\[1.5ex] &= 0.5~\mathrm{mol} \end{align*}\]
NaCl is the limiting reactant.
\[\begin{align*} c(\mathrm{Pb^{2+}}) &= n(\mathrm{Pb^{2+}}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ n(\mathrm{Pb^{2+}})_{\mathrm{initial}} - n(\mathrm{Pb^{2+}})_{\mathrm{reacted}} \biggl\} ~ r(\mathrm{Pb^{2+}, Pb(NO_3)_2}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ \biggr[ c(\mathrm{Pb(NO_3)_2}) ~ V(\mathrm{Pb(NO_3)_2}) \biggr] ~-~ \biggr[ n(\mathrm{PbCl_2}) ~ r(\mathrm{Pb(NO_3)_2,PbCl_2}) \biggr] \biggl\} \\ &\phantom{=}~~~ r(\mathrm{Pb^{2+}, Pb(NO_3)_2}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ \biggr[ \left ( \dfrac{1.0~\mathrm{mol~Pb(NO_3)_2}}{\mathrm{L}} \right ) \left ( \dfrac{1.0~\mathrm{L~Pb(NO_3)_2}}{} \right ) \biggr] - \biggr[ 0.5~\mathrm{mol~PbCl_2} \left ( \dfrac{\mathrm{mol~Pb(NO_3)_2}}{\mathrm{mol~PbCl_2}} \right ) \biggr] \biggl\} \\ &\phantom{=}~~~ \left ( \dfrac{\mathrm{mol~Pb^{2+}}}{\mathrm{PbCl_2}} \right ) \left ( \dfrac{1}{2.0~\mathrm{L}}\right )\\[1.5ex] &= 0.25~\mathrm{mol~L^{-1}}\\[3ex] c(\mathrm{NO_3^{2-}}) &= n(\mathrm{NO_3^{2-}}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= c(\mathrm{Pb(NO_3)_2}) ~ V(\mathrm{Pb(NO_3)_2})_{\mathrm{initial}} ~ r(\mathrm{NO_3^{2-},Pb(NO_3)_2}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \dfrac{1.0~\mathrm{mol~Pb(NO_3)_2}}{\mathrm{L}} \left ( \dfrac{1.0~\mathrm{L}}{} \right ) \left ( \dfrac{2~\mathrm{mol~NO_3^{2-}}}{\mathrm{mol~Pb(NO_3)_2}} \right ) \left ( \dfrac{1}{2.0~\mathrm{L}} \right ) \\[1.5ex] &= 1.0~\mathrm{mol~L^{-1}} \\[3ex] c(\mathrm{Na^+}) &= n(\mathrm{Na^+}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= c(\mathrm{NaCl}) ~ V(\mathrm{NaCl})_{\mathrm{initial}} ~ r(\mathrm{Na^+,NaCl}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \dfrac{1.0~\mathrm{mol~NaCl}}{\mathrm{L}} \left ( \dfrac{1.0~\mathrm{L}}{} \right ) \left ( \dfrac{\mathrm{mol~Na^+}}{\mathrm{mol~NaCl}} \right ) \left ( \dfrac{1}{2.0~\mathrm{L}} \right )\\[1.5ex] &= 0.5~\mathrm{mol~L^{-1}} \end{align*}\]
Because NaCl is the limiting reactant, all Cl– ions are precipitated in PbCl2.
\[\begin{align*} c(\mathrm{Cl^-}) &= 0.0~\mathrm{mol~L^{-1}} \end{align*}\]
Concept: stoichiometry; molarity; dilution; limiting reactant
Determine the molar concentrations (in g mol–1) the ions present in a solution created from mixing equal volumes of 1.0 M aqueous ammonium carbonate and 1.0 M aqueous potassium perchlorate. Assume that the volumes are precise to one decimal place in normalized scientific notation.
Solution
Answer: NH4+: 1.0 M; CO32–: 0.50 M; K+: 0.50 M; ClO4–: 0.50 M
This solution considers the mixing of two 1.0 L solutions to give a 2.0 L final solution.
\[\begin{align*} \mathrm{(NH_4)_2CO_3(aq)} + 2~\mathrm{KClO_4(aq)} \longrightarrow 2~\mathrm{NH_4ClO_4(aq)} + \mathrm{K_2CO_3(aq)} \end{align*}\]
\[\begin{align*} c(\mathrm{NH_4^+}) &= n(\mathrm{NH_4^+}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= n(\mathrm{(NH_4)_2CO_3}) ~ r(\mathrm{NH_4^{+},(NH_4)_2CO_3}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= c(\mathrm{(NH_4)_2CO_3}) ~ V(\mathrm{(NH_4)_2CO_3}) ~ r(\mathrm{NH_4^{+},(NH_4)_2CO_3}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \dfrac{1.0~\mathrm{mol~(NH_4)_2CO_3}}{\mathrm{L}} \left ( \dfrac{1.0~\mathrm{L}}{} \right ) \left ( \dfrac{\mathrm{2~mol~NH_4^{+}}}{\mathrm{mol~(NH_4)_2CO_3}} \right ) \left ( \dfrac{1}{2.0~\mathrm{L}} \right ) \\[1.5ex] &= 1.0~\mathrm{mol~L^{-1}}\\[3ex] c(\mathrm{CO_3^{2-}}) &= n(\mathrm{CO_3^{2-}}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= n(\mathrm{(NH_4)_2CO_3}) ~ r(\mathrm{CO_3^{2-},(NH_4)_2CO_3}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= c(\mathrm{(NH_4)_2CO_3}) ~ V(\mathrm{(NH_4)_2CO_3}) ~ r(\mathrm{CO_3^{2-},(NH_4)_2CO_3}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \dfrac{1.0~\mathrm{mol~(NH_4)_2CO_3}}{\mathrm{L}} \left ( \dfrac{1.0~\mathrm{L}}{} \right ) \left ( \dfrac{\mathrm{mol~CO_3^{2-}}}{\mathrm{mol~(NH_4)_2CO_3}} \right ) \left ( \dfrac{1}{2.0~\mathrm{L}} \right ) \\[1.5ex] &= 0.50~\mathrm{mol~L^{-1}}\\[3ex] c(\mathrm{K^+}) &= n(\mathrm{K^+}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= n(\mathrm{KClO_4}) ~ r(\mathrm{K^+,KClO_4}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= c(\mathrm{KClO_4}) ~ V(\mathrm{KClO_4}) ~ r(\mathrm{K^+,KClO_4}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \dfrac{1.0~\mathrm{mol~KClO_4}}{\mathrm{L}} \left ( \dfrac{1.0~\mathrm{L}}{} \right ) \left ( \dfrac{\mathrm{mol~K^+}}{\mathrm{mol~KClO_4}} \right ) \left ( \dfrac{1}{2.0~\mathrm{L}} \right )\\[1.5ex] &= 0.50~\mathrm{mol~L^{-1}}\\[3ex] c(\mathrm{ClO_4^-}) &= n(\mathrm{ClO_4^-}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= n(\mathrm{KClO_4}) ~ r(\mathrm{ClO_4^-,KClO_4}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= c(\mathrm{KClO_4}) ~ V(\mathrm{KClO_4}) ~ r(\mathrm{ClO_4^-,KClO_4}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \dfrac{1.0~\mathrm{mol~KClO_4}}{\mathrm{L}} \left ( \dfrac{1.0~\mathrm{L}}{} \right ) \left ( \dfrac{\mathrm{mol~ClO_4^-}}{\mathrm{mol~KClO_4}} \right ) \left ( \dfrac{1}{2.0~\mathrm{L}} \right )\\[1.5ex] &= 0.50~\mathrm{mol~L^{-1}} \end{align*}\]
Concept: stoichiometry; molarity; dilution
Determine the molar concentration (in g mol–1) of the salt produced by a reaction between a 200. mL 0.100 M aqueous HCl solution with a 100. mL 0.50 M aqueous KOH solution.
Solution
Answer: 0.0667 M
\[\begin{align*} \mathrm{HCl(aq)} + \mathrm{KOH(aq)} \longrightarrow \mathrm{H_2O(l)} + \mathrm{KCl(aq)} \end{align*}\]
Determine the limiting reactant.
\[\begin{align*} n(\mathrm{KCl}) &= n(\mathrm{HCl}) ~ r(\mathrm{KCl,HCl}) \\[1.5ex] &= c(\mathrm{HCl}) ~ V(\mathrm{initial}) ~ r(\mathrm{KCl,HCl}) \\[1.5ex] &= \dfrac{0.100~\mathrm{mol~HCl}}{\mathrm{L}} \left ( \dfrac{200.~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \left ( \dfrac{\mathrm{mol~KCl}}{\mathrm{mol~HCl}} \right ) \\[1.5ex] &= 0.0200~\mathrm{mol} \\[3ex] n(\mathrm{KCl}) &= n(\mathrm{KOH}) ~ r(\mathrm{KCl,KOH}) \\[1.5ex] &= c(\mathrm{KOH}) ~ V(\mathrm{initial}) ~ r(\mathrm{KCl,KOH}) \\[1.5ex] &= \dfrac{0.50~\mathrm{mol~KOH}}{\mathrm{L}} \left ( \dfrac{100.~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \left ( \dfrac{\mathrm{mol~KCl}}{\mathrm{mol~KOH}} \right ) \\[1.5ex] &= 0.0500~\mathrm{mol} \end{align*}\]
KCl is the limiting reactant.
\[\begin{align*} c(\mathrm{KCl}) &= n(\mathrm{KCl}) ~ V(\mathrm{final})^{-1} \\[1.5ex] &= 0.0200~\mathrm{mol~KCl} \left ( \dfrac{1}{\left (200. + 100. \right ) ~\mathrm{mL}}{} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 0.0667~\mathrm{mol~L^{-1}} \end{align*}\]
Concept: stoichiometry; molarity; acid-base reaction; limiting reactant
What volume (in mL) of a 0.100 M aqueous HNO3 solution is required to neutralize a 50.0 mL 0.150 M aqueous Ba(OH)2 solution?
Solution
Answer:
\[\begin{align*} \mathrm{2~HNO_3(aq)} + \mathrm{Ba(OH)_2(aq)} \longrightarrow \mathrm{2~H_2O(l)} + \mathrm{Ba(NO_3)_2(aq)} \end{align*}\]
\[\begin{align*} V(\mathrm{HNO_3}) &= n(\mathrm{HNO_3}) ~ c(\mathrm{HNO_3})^{-1} \\[1.5ex] &= n(\mathrm{Ba(OH)_2}) ~ r(\mathrm{HNO_3,Ba(OH)_2}) ~ c(\mathrm{HNO_3})^{-1} \\[1.5ex] &= V(\mathrm{Ba(OH)_2}) ~ c(\mathrm{Ba(OH)_2}) ~ r(\mathrm{HNO_3,Ba(OH)_2}) ~ c(\mathrm{HNO_3})^{-1} \\[1.5ex] &= 50.0~\mathrm{mL}~\mathrm{Ba(OH)_2} \left ( \dfrac{0.150~\mathrm{mol~Ba(OH)_2}}{\mathrm{L}} \right ) \left ( \dfrac{\mathrm{2~mol~HNO_3}}{\mathrm{mol~Ba(OH)_2}} \right ) \left ( \dfrac{\mathrm{L}}{0.100~\mathrm{mol}} \right ) \\[1.5ex] &= 150.~\mathrm{mL} \end{align*}\]
Concept: stoichiometry; molarity; acid-base reaction
Hydrogen cyanide is produced industrally from a reaction between gaseous ammonia, oxygen, and methane.
\[\begin{align*} \mathrm{2~NH_3(g)} + \mathrm{3~O_2(g)} + \mathrm{2~CH_4(g)} \longrightarrow \mathrm{2~HCN(g)} + \mathrm{6~H_2O(l)} \\ \end{align*}\]
If 5.00 × 103 kg of each reactant react, what mass (in kg) of each product would be produced (assuming a 100 % yield)?
Solution
Answer: HCN: 2.82 × 103 kg; H2O: 5.63 × 103 kg
Find the limiting reactant.
\[\begin{align*} n(\mathrm{HCN}) &= n(\mathrm{NH_3}) ~ r(\mathrm{HCN,NH_3}) \\[1.5ex] &= m(\mathrm{NH_3}) ~ M(\mathrm{NH_3})^{-1} ~ r(\mathrm{HCN,NH_3}) \\[1.5ex] &= 5.00 \times 10^{3}~\mathrm{kg~NH_3} \left ( \dfrac{10^3~\mathrm{g}}{\mathrm{kg}} \right ) \left ( \dfrac{\mathrm{mol~NH_3}}{\mathrm{17.04~g}} \right ) \left ( \dfrac{\mathrm{2~mol~HCN}}{\mathrm{2~mol~NH_3}} \right )\\[1.5ex] &= 2.9\bar{3}42\times 10^{5}~\mathrm{mol} \\[1.0ex] n(\mathrm{HCN}) &= n(\mathrm{O_2}) ~ r(\mathrm{HCN,O_2}) \\[1.5ex] &= m(\mathrm{O_2}) ~ M(\mathrm{O_2})^{-1} ~ r(\mathrm{HCN,O_2}) \\[1.5ex] &= 5.00 \times 10^{3}~\mathrm{kg~O_2} \left ( \dfrac{10^3~\mathrm{g}}{\mathrm{kg}} \right ) \left ( \dfrac{\mathrm{mol~O_2}}{\mathrm{32.00~g}} \right ) \left ( \dfrac{\mathrm{2~mol~HCN}}{\mathrm{3~mol~O_2}} \right )\\[1.5ex] &= 1.0\bar{4}16\times 10^{5}~\mathrm{mol} \\[3ex] n(\mathrm{HCN}) &= n(\mathrm{CH_4}) ~ r(\mathrm{HCN,CH_4}) \\[1.5ex] &= m(\mathrm{CH_4}) ~ M(\mathrm{CH_4})^{-1} ~ r(\mathrm{HCN,CH_4}) \\[1.5ex] &= 5.00 \times 10^{3}~\mathrm{kg~CH_4} \left ( \dfrac{10^3~\mathrm{g}}{\mathrm{kg}} \right ) \left ( \dfrac{\mathrm{mol~CH_4}}{\mathrm{16.05~g}} \right ) \left ( \dfrac{\mathrm{2~mol~HCN}}{\mathrm{2~mol~CH_4}} \right )\\[1.5ex] &= 3.1\bar{1}52\times 10^{5}~\mathrm{mol} \end{align*}\]
O2 is limiting.
\[\begin{align*} m(\mathrm{HCN}) &= n(\mathrm{HCN}) ~ M(\mathrm{HCN}) \\[1.5ex] &= 1.0\bar{4}16\times 10^{5}~\mathrm{mol~HCN} \left ( \dfrac{27.03~\mathrm{g}}{\mathrm{mol}} \right ) \left ( \dfrac{\mathrm{kg}}{10^3~\mathrm{g}} \right ) \\[1.5ex] &= 2.8\bar{1}54\times 10^{3}~\mathrm{kg} = 2.82\times 10^{3}~\mathrm{kg}\\[3ex] m(\mathrm{H_2O}) &= n(\mathrm{H_2O}) ~ M(\mathrm{H_2O}) \\[1.5ex] &= n(\mathrm{HCN}) ~ r(\mathrm{H_2O,HCN}) ~ M(\mathrm{H_2O}) \\[1.5ex] &= 1.0\bar{4}16\times 10^{5}~\mathrm{mol~HCN} \left ( \dfrac{6~\mathrm{mol~H_2O}}{2~\mathrm{mol~HCN}} \right ) \left ( \dfrac{18.02~\mathrm{g}}{\mathrm{mol}} \right ) \left ( \dfrac{\mathrm{kg}}{10^3~\mathrm{g}} \right ) \\[1.5ex] &= 5.6\bar{3}08\times 10^{3}~\mathrm{kg} = 5.63\times 10^{3}~\mathrm{kg} \end{align*}\]
Concept: stoichiometry; molarity; dilution; limiting reactant
Acrylonitrile (C3H3N) is the starting material for many synthetic carpets and fabrics and is produced by the following reaction.
\[\begin{align*} \mathrm{2~C_3H_6(g)} + \mathrm{2~NH_3(g)} + \mathrm{3~O_2(g)} \longrightarrow \mathrm{2~C_3H_3N(g)} + \mathrm{6~H_2O(g)} \\ \end{align*}\]
If 15.0 g C3H6, 5.00 g NH3, and 8.00 g O2 react, what mass (in g) of acrylonitrile can be produced (assuming a 100 % yield)?
Solution
Answer: 11.1 g
Find the limiting reactant.
\[\begin{align*} n(\mathrm{C_3H_3N}) &= n(\mathrm{C_3H_6}) ~ r(\mathrm{C_3H_3N,C_3H_6}) \\[1.5ex] &= m(\mathrm{C_3H_6}) ~ M(\mathrm{C_3H_6})^{-1} ~ r(\mathrm{C_3H_3N,C_3H_6}) \\[1.5ex] &= 15.0~\mathrm{g~C_3H_6} \left ( \dfrac{\mathrm{mol~C_3H_6}}{\mathrm{53.07~g}} \right ) \left ( \dfrac{\mathrm{2~mol~C_3H_3N}}{\mathrm{2~mol~C_3H_6}} \right )\\[1.5ex] &= 0.28\bar{2}64~\mathrm{mol} \\[3ex] n(\mathrm{C_3H_3N}) &= n(\mathrm{NH_3}) ~ r(\mathrm{C_3H_3N,NH_3}) \\[1.5ex] &= m(\mathrm{NH_3}) ~ M(\mathrm{NH_3})^{-1} ~ r(\mathrm{C_3H_3N,NH_3}) \\[1.5ex] &= 5.00~\mathrm{g~NH_3} \left ( \dfrac{\mathrm{mol~NH_3}}{\mathrm{17.04~g}} \right ) \left ( \dfrac{\mathrm{2~mol~C_3H_3N}}{\mathrm{2~mol~NH_3}} \right )\\[1.5ex] &= 0.29\bar{3}42~\mathrm{mol} \\[3ex] n(\mathrm{C_3H_3N}) &= n(\mathrm{O_2}) ~ r(\mathrm{C_3H_3N,O_2}) \\[1.5ex] &= m(\mathrm{O_2}) ~ M(\mathrm{O_2})^{-1} ~ r(\mathrm{C_3H_3N,O_2}) \\[1.5ex] &= 8.00~\mathrm{g~O_2} \left ( \dfrac{\mathrm{mol~O_2}}{\mathrm{32.00~g}} \right ) \left ( \dfrac{\mathrm{2~mol~C_3H_3N}}{\mathrm{3~mol~O_2}} \right )\\[1.5ex] &= 0.16\bar{6}66~\mathrm{mol} \end{align*}\]
O2 is limiting.
\[\begin{align*} m(\mathrm{C_3H_3N}) &= n(\mathrm{C_3H_3N}) ~ M(\mathrm{C_3H_3N}) \\[1.5ex] &= 0.16\bar{6}66~\mathrm{mol}~\mathrm{mol~C_3H_3N} \left ( \dfrac{6~\mathrm{mol~NH_3}}{2~\mathrm{mol~C_3H_3N}} \right ) \left ( \dfrac{18.02~\mathrm{g}}{\mathrm{mol}} \right ) \\[1.5ex] &= 8.8\bar{4}46~\mathrm{g} = 8.84~\mathrm{g} \end{align*}\]
Concept: stoichiometry; limiting reactant
Calcium chloride is a strong electrolyte and is used to ``salt’’ streets in the winter to melt ice and snow. Write a net ionic reaction to show how this substance breaks apart when it dissolves in water.
Solution
Answer:
\[\begin{align*} \mathrm{CaCl_2(s)} \longrightarrow \mathrm{Ca^{2+}(aq)} + \mathrm{2~Cl^-(aq)} \end{align*}\]
Concept: net ionic equation
A solution of ethanol in water is prepared by dissolving 75.0 mL of ethanol (ρ = 0.79 g cm–3) in enough water to make a 250.0 mL solution. What is the molar concentration (in mol L–1) of the ethanol in this solution?
Solution
Answer: 5.14 M
\[\begin{align*} c(\mathrm{C_2H_6O}) &= n(\mathrm{C_2H_6O}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= m(\mathrm{C_2H_6O}) ~ M(\mathrm{C_2H_6O})^{-1} ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= V(\mathrm{C_2H_6O}) ~ \rho(\mathrm{C_2H_6O}) ~ M(\mathrm{C_2H_6O})^{-1} ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= 75.0~\mathrm{mL} \left ( \dfrac{\mathrm{cm^3}}{\mathrm{mL}} \right ) \left ( \dfrac{\mathrm{0.79~g}}{\mathrm{cm^3}} \right ) \left ( \dfrac{\mathrm{mol~C_2H_6O}}{46.08~\mathrm{g}} \right ) \left ( \dfrac{1}{250.0~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 5.1\bar{4}32~\mathrm{mol~L^{-1}} = 5.14~\mathrm{mol~L^{-1}} \end{align*}\]
Concept: stoichiometry; molarity
Which of the following aqueous solutions contains the largest number of ions?
- 100.0 mL of 0.100 M NaOH
- 50.0 mL of 0.200 M BaCl2
- 75.0 mL of 0.150 M Na3PO4
Solution
Answer: C
A
\[\begin{align*} \mathrm{NaOH(aq)} \longrightarrow \mathrm{Na^+(aq)} + \mathrm{OH^-(aq)} \end{align*}\]
\[\begin{align*} n(\mathrm{ions}) &= n(\mathrm{NaOH}) ~ r(\mathrm{ions,NaOH}) \\[1.5ex] &= c(\mathrm{NaOH}) ~ V(\mathrm{NaOH}) ~ r(\mathrm{ions,NaOH}) \\[1.5ex] &= \dfrac{0.100~\mathrm{mol}}{\mathrm{L}} \left ( \dfrac{100.0~\mathrm{mL}}{1} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \left ( \dfrac{2~\mathrm{mol~ions}}{\mathrm{mol~NaOH}} \right ) \\[1.5ex] &= 0.0200~\mathrm{mol} \end{align*}\]
B
\[\begin{align*} \mathrm{BaCl_2(aq)} \longrightarrow \mathrm{Ba^{2+}(aq)} + 2~\mathrm{Cl^-(aq)} \end{align*}\]
\[\begin{align*} n(\mathrm{ions}) &= n(\mathrm{BaCl_2}) ~ r(\mathrm{ions,BaCl_2}) \\[1.5ex] &= c(\mathrm{BaCl_2}) ~ V(\mathrm{BaCl_2}) ~ r(\mathrm{ions,BaCl_2}) \\[1.5ex] &= \dfrac{0.200~\mathrm{mol}}{\mathrm{L}} \left ( \dfrac{50.0~\mathrm{mL}}{1} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \left ( \dfrac{3~\mathrm{mol~ions}}{\mathrm{mol~BaCl_2}} \right ) \\[1.5ex] &= 0.0300~\mathrm{mol} \end{align*}\]
C
\[\begin{align*} \mathrm{Na_3PO_4(aq)} \longrightarrow \mathrm{3~Na^2+(aq)} + \mathrm{PO_4^{3-}(aq)} \end{align*}\]
\[\begin{align*} n(\mathrm{ions}) &= n(\mathrm{Na_3PO_4}) ~ r(\mathrm{ions,Na_3PO_4}) \\[1.5ex] &= c(\mathrm{Na_3PO_4}) ~ V(\mathrm{Na_3PO_4}) ~ r(\mathrm{ions,Na_3PO_4}) \\[1.5ex] &= \dfrac{0.150~\mathrm{mol}}{\mathrm{L}} \left ( \dfrac{75.0~\mathrm{mL}}{1} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \left ( \dfrac{4~\mathrm{mol~ions}}{\mathrm{mol~Na_3PO_4}} \right ) \\[1.5ex] &= 0.0450~\mathrm{mol} \end{align*}\]
C has the largest number of ions.
Concept: stoichiometry; molarity; ions
If 12.0 g of AgNO3 is available, what volume (in L) of 0.25 M AgNO3 can be prepared?
Solution
Answer: 0.28 L
\[\begin{align*} V(\mathrm{AgNO_3}) &= n(\mathrm{AgNO_3}) ~ c(\mathrm{AgNO_3})^{-1} \\[1.5ex] &= m(\mathrm{AgNO_3}) ~ M(\mathrm{AgNO_3})^{-1} ~ c(\mathrm{AgNO_3})^{-1} \\[1.5ex] &= 12.0~\mathrm{g~AgNO_3} \left ( \dfrac{\mathrm{mol~AgNO_3}}{169.88~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{L}}{\mathrm{0.25~mol~AgNO_3}} \right ) \\[1.5ex] &= 0.2\bar{8}25~\mathrm{L} = 0.28~\mathrm{L} \end{align*}\]
Concept: stoichiometry; molarity
A solution is prepared by dissolving 10.8 g ammonium sulfate in enough water to make 100.0 mL of stock solution. A 10.00 mL aliquot is taken and 50.00 mL of water is added. What is the molar concentration (in mol L–1) of ammonium ions and sulfate ions in the final solution?
Solution
Answer: NH4+: 0.272 M; SO42–: 0.136 M
\[\begin{align*} c(\mathrm{NH_4^+}) &= n(\mathrm{NH_4^+})_{\textrm{dilute aliquot}} ~ V(\mathrm{dilute~aliquot})^{-1} \\[1.5ex] &= n(\mathrm{NH_4^+})_{\textrm{stock solution}} ~ V(\mathrm{stock~solution})^{-1} ~ V(\mathrm{aliquot}) ~ V(\mathrm{dilute~aliquot})^{-1} \\[1.5ex] &= m(\mathrm{(NH_4)_2SO_4}) ~ M(\mathrm{(NH_4)_2SO_4})^{-1} ~ r(\mathrm{NH_4^+,(NH_4)_2SO_4})\\ &\phantom{=}~~ V(\mathrm{stock~solution})^{-1} ~ V(\mathrm{aliquot}) ~ V(\mathrm{dilute~aliquot})^{-1} \\[1.5ex] &= 10.8~\mathrm{g~(NH_4)_2SO_4} \left ( \dfrac{\mathrm{mol~(NH_4)_2SO_4}}{132.16~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{2~mol~NH_4^+}}{\mathrm{mol~(NH_4)_2SO_4}} \right ) \\ &\phantom{=(} \left ( \dfrac{1}{100.0~\mathrm{mL}} \right ) \left ( \dfrac{10.0~\mathrm{mL}}{} \right ) \left ( \dfrac{1}{60.0~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 0.27\bar{2}39~\mathrm{mol~L^{-1}} = 0.272~\mathrm{mol~L^{-1}} \\[3ex] c(\mathrm{SO_4^{2-}}) &= n(\mathrm{SO_4^{2-}})_{\textrm{dilute~aliquot}} ~ V(\mathrm{stock~solution})^{-1} \\[1.5ex] &= n(\mathrm{SO_4^{2-}})_{\textrm{stock~solution}} ~ V(\mathrm{stock~solution})^{-1}~ V(\mathrm{aliquot}) ~ V(\mathrm{dilute~aliquot})^{-1} \\[1.5ex] &= m(\mathrm{(NH_4)_2SO_4}) ~ M(\mathrm{(NH_4)_2SO_4})^{-1} ~ r(\mathrm{SO_4^{2-},(NH_4)_2SO_4})\\ &\phantom{=}~~ V(\mathrm{stock~solution})^{-1} ~ V(\mathrm{aliquot}) ~ V(\mathrm{dilute~aliquot})^{-1} \\[1.5ex] &= 10.8~\mathrm{g~(NH_4)_2SO_4} \left ( \dfrac{\mathrm{mol~(NH_4)_2SO_4}}{132.16~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{mol~SO_4^{2-}}}{\mathrm{mol~(NH_4)_2SO_4}} \right ) \\ &\phantom{=(} \left ( \dfrac{1}{100.0~\mathrm{mL}} \right ) \left ( \dfrac{10.0~\mathrm{mL}}{} \right ) \left ( \dfrac{1}{60.0~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 0.13\bar{6}19~\mathrm{mol~L^{-1}} = 0.136~\mathrm{mol~L^{-1}} \end{align*}\]
Alternatively,
\[\begin{align*} c(\mathrm{SO_4^{2-}}) &= c(\mathrm{NH_4^+})_{\textrm{dilute~aliquot}} ~ r(\mathrm{SO_4^{2-},NH_4^+}) \\[1.5ex] &= \dfrac{0.27\bar{2}39~\mathrm{mol}}{\mathrm{L}} \left ( \dfrac{\mathrm{mol~SO_4^{2-}}}{\mathrm{2~mol~NH_4^+}} \right ) \\[1.5ex] &= 0.13\bar{6}19~\mathrm{mol~L^{-1}} = 0.136~\mathrm{mol~L^{-1}} \end{align*}\]
Concept: stoichiometry; molarity; dilution
What mass (in g) of Na2CrO4 is required to precipitate all of the silver ions from a 75.0 mL 0.100 M aqueous solution of AgNO3?
Solution
Answer: 0.607 g
\[\begin{align*} \mathrm{Na_2CrO_4(aq)} + 2~\mathrm{AgNO_3(aq)} \longrightarrow \mathrm{2~NaNO_3(aq)} + \mathrm{Ag_2CrO_4}(s) \end{align*}\]
\[\begin{align*} m(\mathrm{Na_2CrO_4}) &= n(\mathrm{Na_2CrO_4}) ~ M(\mathrm{Na_2CrO_4}) \\[1.5ex] &= n(\mathrm{AgNO_3}) ~ r(\mathrm{Na_2CrO_4,AgNO_3}) ~ M(\mathrm{Na_2CrO_4}) \\[1.5ex] &= c(\mathrm{AgNO_3}) ~ V(\mathrm{AgNO_3}) ~ r(\mathrm{Na_2CrO_4,AgNO_3}) ~ M(\mathrm{Na_2CrO_4}) \\[1.5ex] &= \dfrac{0.100~\mathrm{mol~AgNO_3}}{\mathrm{L}} \left ( \dfrac{75.0~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \left ( \dfrac{\mathrm{mol~Na_2CrO_4}}{\mathrm{2~mol~AgNO_3}} \right ) \left ( \dfrac{161.98~\mathrm{g~Na_2CrO_4}}{\mathrm{mol}} \right )\\[1.5ex] &= 0.60\bar{7}42~\mathrm{g} = 0.607~\mathrm{g} \end{align*}\]
Concept: stoichiometry; molarity; precipitation
What mass (in g) of iron(III) hydroxide precipitate can be produced by reacting a 72.0 mL 0.105 M aqueous iron(III) nitrate solution with a 125 mL 0.150 M aqueous sodium hydroxide solution?
Solution
Answer: 0.668 g
\[\begin{align*} \mathrm{Fe(NO_3)_3(aq)} + 3~\mathrm{NaOH(aq)} \longrightarrow \mathrm{Fe(OH)_3(s)} + \mathrm{3~NaNO_3(aq)} \end{align*}\]
Determine the limiting reactant.
\[\begin{align*} n(\mathrm{Fe(OH)_3}) &= n(\mathrm{Fe(NO_3)_3}) ~ r(\mathrm{Fe(OH)_3,Fe(NO_3)_3}) \\[1.5ex] &= c(\mathrm{Fe(NO_3)_3}) ~ V(\mathrm{initial}) ~ r(\mathrm{Fe(OH)_3,Fe(NO_3)_3}) \\[1.5ex] &= \dfrac{0.105~\mathrm{mol~Fe(NO_3)_3}}{\mathrm{L}} \left ( \dfrac{72.0~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \left ( \dfrac{\mathrm{mol~Fe(OH)_3}}{\mathrm{mol~Fe(NO_3)_3}} \right ) \\[1.5ex] &= 7.5\bar{6}00 \times 10^{-3}~\mathrm{mol}\\[3ex] n(\mathrm{Fe(OH)_3}) &= n(\mathrm{NaOH}) ~ r(\mathrm{Fe(OH)_3,NaOH}) \\[1.5ex] &= c(\mathrm{NaOH}) ~ V(\mathrm{initial}) ~ r(\mathrm{Fe(OH)_3,NaOH}) \\[1.5ex] &= \dfrac{0.150~\mathrm{mol~NaOH}}{\mathrm{L}} \left ( \dfrac{125~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \left ( \dfrac{\mathrm{mol~Fe(OH)_3}}{\mathrm{3~mol~NaOH}} \right ) \\[1.5ex] &= 6.2\bar{5}00 \times 10^{-3}~\mathrm{mol} \end{align*}\]
NaOH is limiting.
\[\begin{align*} m(\mathrm{Fe(OH)_3}) &= n(\mathrm{Fe(OH)_3}) ~ M(\mathrm{Fe(OH)_3}) \\[1.5ex] &= 6.2\bar{5}00 \times 10^{-3}~\mathrm{mol~Fe(OH)_3} \left ( \dfrac{106.88~\mathrm{g~Fe(OH)_3}}{\mathrm{mol}} \right )\\[1.5ex] &= 0.66\bar{8}00~\mathrm{g} = 0.668~\mathrm{g} \end{align*}\]
Concept: stoichiometry; molarity; precipitation; limiting reactant
A 100.0 mL 0.200 M aqueous potassium hydroxide solution is mixed with a 100.0 mL 0.200 M aqueous magnesium nitrate solution.
- Write a balanced chemical equation for the reaction that occurs.
- Determine the precipitate that forms (if any).
- Determine the mass (in g) of precipitate that forms (if any).
- Determine the molar concentration (in g mol–1) of each ion in solution after the reaction goes to 100 % completion.
Solution
Answer:
A
\[\begin{align*} \mathrm{2~KOH(aq)} + \mathrm{Mg(NO_3)_2(aq)} \longrightarrow \mathrm{2~KNO_3(aq)} + \mathrm{Mg(OH)_2(s)} \end{align*}\]
B
Mg(OH)2
C
Determine the limiting reactant.
\[\begin{align*} n(\mathrm{Mg(OH)_2}) &= c(\mathrm{KOH}) ~ V(\mathrm{initial}) ~ r(\mathrm{Mg(OH)_2,KOH}) \\[1.5ex] &= \dfrac{0.200~\mathrm{mol~KOH}}{\mathrm{L}} \left ( \dfrac{100.0~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \left ( \dfrac{\mathrm{mol~Mg(OH)_2}}{\mathrm{2~mol~KOH}} \right ) \\[1.5ex] &= 0.0100~\mathrm{mol}\\[3ex] n(\mathrm{Mg(OH)_2}) &= c(\mathrm{Mg(NO_3)_2}) ~ V(\mathrm{initial}) ~ r(\mathrm{Mg(OH)_2,Mg(NO_3)_2}) \\[1.5ex] &= \dfrac{0.200~\mathrm{mol~Mg(NO_3)_2}}{\mathrm{L}} \left ( \dfrac{100.0~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \left ( \dfrac{\mathrm{mol~Mg(OH)_2}}{\mathrm{mol~Mg(NO_3)_2}} \right ) \\[1.5ex] &= 0.0200~\mathrm{mol} \end{align*}\]
KOH is limiting.
\[\begin{align*} m(\mathrm{Mg(OH)_2}) &= n(\mathrm{Mg(OH)_2}) ~ M(\mathrm{Mg(OH)_2}) \\[1.5ex] &= 0.0100~\mathrm{mol~Mg(OH)_2} \left ( \dfrac{58.33~\mathrm{g~Mg(OH)_2}}{\mathrm{mol}} \right )\\[1.5ex] &= 0.58\bar{3}30~\mathrm{g} = 0.583~\mathrm{g} \end{align*}\]
D
\[\begin{align*} c(\mathrm{K^+}) &= n(\mathrm{K^+}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= n(\mathrm{KOH}) ~ r(\mathrm{K^+,KOH}) V(\mathrm{solution})^{-1} \\[1.5ex] &= c(\mathrm{KOH}) ~ V(\mathrm{KOH}) ~ r(\mathrm{K^+,KOH}) V(\mathrm{solution})^{-1} \\[1.5ex] &= \dfrac{0.200~\mathrm{mol~KOH}}{\mathrm{L}} \left ( \dfrac{100.0~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{mol~K^+}}{\mathrm{mol~KOH}} \right ) \left ( \dfrac{1}{(100.0 + 100.0)~\mathrm{mL}} \right ) \\[1.5ex] &= 0.100~\mathrm{mol~L^{-1}} \\[3ex] c(\mathrm{OH^-}) &= 0 ~\textrm{since KOH is limiting and all hydroxide precipitates as}~\mathrm{Mg(OH)_2} \\[3ex] c(\mathrm{Mg^{2+}})_{\mathrm{final}} &= n(\mathrm{Mg^{2+}}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ n(\mathrm{Mg(NO_3)_2})_{\mathrm{initial}} - n(\mathrm{Mg(NO_3)_2})_{\mathrm{reacted}} \biggl\} ~ r(\mathrm{Mg^{2+},Mg(NO_3)_2}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \Biggl\{ \biggr[ c(\mathrm{Mg(NO_3)_2}) ~ V(\mathrm{Mg(NO_3)_2}) \biggr] - \biggr[ n(\mathrm{Mg(OH)_2})_{\mathrm{formed}}~ r(\mathrm{Mg(NO_3)_2,Mg(OH)_2}) \biggr] \Biggl\} \\ &\phantom{=}~~~ r(\mathrm{Mg^{2+},Mg(NO_3)_2}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \Biggl\{\biggr[ \left ( \dfrac{0.200~\mathrm{mol~Mg(NO_3)_2}}{\mathrm{L}} \right ) \left ( \dfrac{100.0~\mathrm{mL~Mg(NO_3)_2}}{} \right ) \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \biggr] ~- &\phantom{=\Biggl\{~} ~\biggr[ \left ( \dfrac{0.0100~\mathrm{mol~Mg(OH)_2}}{} \right ) \left ( \dfrac{\mathrm{mol~Mg(NO_3)_2}}{\mathrm{mol~Mg(OH)_2}} \right ) \biggr] \Biggl\}\\ &\phantom{=\Biggl\{}~ \left ( \dfrac{\mathrm{mol~Mg^{2+}}}{\mathrm{mol~Mg(NO_3)_2}} \right ) \left ( \dfrac{1}{(100.0 + 100.0)~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 0.0500~\mathrm{mol~L^{-1}} \\[3ex] c(\mathrm{NO_3^-}) &= n(\mathrm{NO_3^-}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= n(\mathrm{Mg(NO_3)_2}) ~ r(\mathrm{NO_3^-,Mg(NO_3)_2}) V(\mathrm{solution})^{-1} \\[1.5ex] &= c(\mathrm{Mg(NO_3)_2}) ~ V(\mathrm{Mg(NO_3)_2}) ~ r(\mathrm{NO_3^-,Mg(NO_3)_2}) V(\mathrm{solution})^{-1} \\[1.5ex] &= \dfrac{0.200~\mathrm{mol~Mg(NO_3)_2}}{\mathrm{L}} \left ( \dfrac{100.0~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{2~mol~NO_3^-}}{\mathrm{mol~Mg(NO_3)_2}} \right ) \left ( \dfrac{1}{(100.0 + 100.0)~\mathrm{mL}} \right ) \\[1.5ex] &= 0.200~\mathrm{mol~L^{-1}} \end{align*}\]
Concept: balancing equations; stoichiometry; molarity; precipitation; limiting reactant
A 1.42 g sample of a pure, metal (M) containing compound (M2SO4) was dissolved in water and treated with an excess of aqueous calcium chloride. All the sulfate ions precipitated as calcium sulfate which was collected, dried, and found to be 1.36 g. What is the identity and standard atomic weight of the metal?
Solution
Answer: Na; 23.0
\[\begin{align*} \mathrm{M_2SO_4(aq)} + \mathrm{CaCl_2(aq)} \longrightarrow \mathrm{CaSO_4(s)} + \mathrm{2~MCl(aq)} \end{align*}\]
\[\begin{align*} M(\mathrm{M^+}) &= m(\mathrm{M^+}) ~ n(\mathrm{M^+})^{-1} \\[1.5ex] &= \biggl\{ m(\mathrm{M_2SO_4}) - m(\mathrm{SO_4^{2-}}) \biggl\} ~ \biggl\{ n(\mathrm{M_2SO_4}) ~ r(\mathrm{M_2SO_4,SO_4^{2-}}) ~ r(\mathrm{M^+,M_2SO_4}) \biggl\}^{-1} ~M_{\mathrm{u}} \\[1.5ex] &= \biggl\{ m(\mathrm{M_2SO_4}) - m(\mathrm{SO_4^{2-}}) \biggl\} ~ \biggl\{ n(\mathrm{CaSO_4}) ~ r(\mathrm{SO_4^{2-},CaSO_4}) ~ r(\mathrm{M_2SO_4,SO_4^{2-}}) ~ r(\mathrm{M^+,M_2SO_4}) \biggl\}^{-1} ~M_{\mathrm{u}} \\[1.5ex] &= \biggl\{ m(\mathrm{M_2SO_4}) - \biggr[ m(\mathrm{CaSO_4}) ~ M(\mathrm{CaSO_4})^{-1} ~ r(\mathrm{SO_4^{2-},CaSO_4}) \biggr] \biggl\} ~ \\[1.5ex] &\phantom{=~~~} \biggl\{ m(\mathrm{CaSO_4}) ~ M(\mathrm{CaSO_4})^{-1} ~ r(\mathrm{M_2SO_4,CaSO_4}) ~ r(\mathrm{M_2SO_4,SO_4^{2-}}) ~ r(\mathrm{M^+,M_2SO_4}) \biggl\}^{-1} ~ M_{\mathrm{u}} \\[1.5ex] &= \Biggl\{ 1.42~\mathrm{g~M_2SO_4} - \Biggr[ 1.36~\mathrm{g~CaSO_4} \left ( \dfrac{\mathrm{mol~CaSO_4}}{136.14~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{mol~SO_4^{2-}}}{\mathrm{mol~CaSO_4}} \right ) \left ( \dfrac{96.06~\mathrm{g}}{\mathrm{mol~SO_4^{2-}}} \right ) \Biggr] \Biggl\} \\[1.5ex] &\phantom{=~~~} \Biggl\{ 1.36~\mathrm{g~CaSO_4} \left ( \dfrac{\mathrm{mol~CaSO_4}}{136.14~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{mol~SO_4^{2-}}}{\mathrm{mol~CaSO_4}} \right ) \left ( \dfrac{\mathrm{mol~M_2SO_4}}{\mathrm{mol~SO_4^{2-}}} \right ) \left ( \dfrac{\mathrm{2~mol~M^+}}{\mathrm{mol~M_2SO_4}} \right ) \Biggl\}^{-1} \left ( \dfrac{\mathrm{mol}}{\mathrm{g}} \right ) \\[1.5ex] &= \dfrac{0.46\bar{0}38~\mathrm{g}}{0.019\bar{9}79~\mathrm{mol}} \left ( \dfrac{\mathrm{mol}}{\mathrm{g}}\right ) \\[1.5ex] &= 23.\bar{0}43 = 23.0 \end{align*}\]
The metal is sodium (Na).
Concept: stoichiometry; molarity; precipitation; molar mass
What volume (in mL) of each of the following bases will will completely react with 25.0 mL of 0.200 M HCl?
- 0.100 M NaOH
- 0.0500 M Sr(OH)2
- 0.250 M KOH
Solution
Answer: 50.0 mL; 50.0 mL; 20.0 mL
A
\[\begin{align*} \mathrm{HCl(aq)} + \mathrm{NaOH(aq)} \longrightarrow \mathrm{H_2O(l)} + \mathrm{NaCl(aq)} \end{align*}\]
\[\begin{align*} V(\mathrm{NaOH}) &= n(\mathrm{NaOH}) ~ c(\mathrm{NaOH})^{-1} \\[1.5ex] &= n(\mathrm{HCl}) ~ r(\mathrm{NaOH,HCl}) ~ c(\mathrm{NaOH})^{-1} \\[1.5ex] &= c(\mathrm{HCl}) ~ V(\mathrm{HCl}) ~ r(\mathrm{NaOH,HCl}) ~ c(\mathrm{NaOH})^{-1} \\[1.5ex] &= \dfrac{0.200~\mathrm{mol~NaOH}}{\mathrm{L}} \left ( \dfrac{25.0~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{mol~NaOH}}{\mathrm{mol~HCl}} \right ) \left ( \dfrac{\mathrm{L}}{0.100~\mathrm{mol~NaOH}} \right ) \\[1.5ex] &= 50.0~\mathrm{mL} \end{align*}\]
B
\[\begin{align*} \mathrm{2~HCl(aq)} + \mathrm{Sr(OH)_2(aq)} \longrightarrow \mathrm{2~H_2O(l)} + \mathrm{SrCl_2(aq)} \end{align*}\]
\[\begin{align*} V(\mathrm{Sr(OH)_2}) &= n(\mathrm{Sr(OH)_2}) ~ c(\mathrm{Sr(OH)_2})^{-1} \\[1.5ex] &= n(\mathrm{HCl}) ~ r(\mathrm{Sr(OH)_2,HCl}) ~ c(\mathrm{Sr(OH)_2})^{-1} \\[1.5ex] &= c(\mathrm{HCl}) ~ V(\mathrm{HCl}) ~ r(\mathrm{Sr(OH)_2,HCl}) ~ c(\mathrm{Sr(OH)_2})^{-1} \\[1.5ex] &= \dfrac{0.200~\mathrm{mol~NaOH}}{\mathrm{L}} \left ( \dfrac{25.0~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{mol~Sr(OH)_2}}{\mathrm{2~mol~HCl}} \right ) \left ( \dfrac{\mathrm{L}}{0.0500~\mathrm{mol~Sr(OH)_2}} \right ) \\[1.5ex] &= 50.0~\mathrm{mL} \end{align*}\]
C
\[\begin{align*} \mathrm{HCl(aq)} + \mathrm{KOH(aq)} \longrightarrow \mathrm{H_2O(l)} + \mathrm{KCl(aq)} \end{align*}\]
\[\begin{align*} V(\mathrm{KOH}) &= n(\mathrm{KOH}) ~ c(\mathrm{KOH})^{-1} \\[1.5ex] &= n(\mathrm{HCl}) ~ r(\mathrm{KOH,HCl}) ~ c(\mathrm{KOH})^{-1} \\[1.5ex] &= c(\mathrm{HCl}) ~ V(\mathrm{HCl}) ~ r(\mathrm{KOH,HCl}) ~ c(\mathrm{KOH})^{-1} \\[1.5ex] &= \dfrac{0.200~\mathrm{mol~NaOH}}{\mathrm{L}} \left ( \dfrac{25.0~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{mol~KOH}}{\mathrm{mol~HCl}} \right ) \left ( \dfrac{\mathrm{L}}{0.250~\mathrm{mol~KOH}} \right ) \\[1.5ex] &= 20.0~\mathrm{mL} \end{align*}\]
Concept: balancing equations; stoichiometry; molarity; acid-base reaction
A 25.0 mL sample of HCl(aq) requires 24.16 mL of 0.106 M NaOH for complete neutralization. What is the molar concentration (in mol L–1) of the original HCl(aq) solution?
Solution
Answer:
\[\begin{align*} \mathrm{HCl(aq)} + \mathrm{NaOH(aq)} \longrightarrow \mathrm{H_2O(l)} + \mathrm{NaCl(aq)} \end{align*}\]
\[\begin{align*} c(\mathrm{HCl}) &= n(\mathrm{HCl}) ~ V(\mathrm{HCl})^{-1} \\[1.5ex] &= n(\mathrm{NaOH}) ~ r(\mathrm{HCl,NaOH})~ V(\mathrm{HCl})^{-1} \\[1.5ex] &= c(\mathrm{NaOH}) ~ V(\mathrm{NaOH}) ~ r(\mathrm{HCl,NaOH})~ V(\mathrm{HCl})^{-1} \\[1.5ex] &= \dfrac{0.106~\mathrm{mol~NaOH}}{\mathrm{L}} \left ( \dfrac{24.16~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \left ( \dfrac{\mathrm{mol~HCl}}{\mathrm{mol~NaOH}} \right ) \left ( \dfrac{1}{25.0~\mathrm{mL~HCl}} \right ) \\[1.5ex] &= 0.102~\mathrm{mol~L^{-1}} \end{align*}\]
Concept: balancing equations; stoichiometry; molarity; acid-base reaction
5.00 g of barium chloride was added to 225 mL of a 1.40 M solution of sodium sulfide.
- Write the balanced molecular equation and include phase labels.
- Write the full ionic equation and include phase labels.
- Write the net ionic equation and include phase labels. If there is no net ionic equation, write “no net ionic equation.”
- Indicate the pricipitate (if any).
- Determine the limiting reactant.
Solution
Answer:
A
Molecular equation
\[\begin{align*} \mathrm{BaCl_2(aq)} + \mathrm{Na_2S(aq)} &\longrightarrow \mathrm{BaS(aq)} + 2~\mathrm{NaCl(aq)}\\ \end{align*}\]
B
Complete ionic equation
\[\begin{align*} \mathrm{Ba^{2+}(aq)} + 2~\mathrm{Cl^-(aq)} + 2~\mathrm{Na^+(aq)} + \mathrm{S^{2-}(aq)} &\longrightarrow \mathrm{Ba^{2+}(aq)} + \mathrm{S^{2-}(aq)} + 2~\mathrm{Na^+(aq)} + 2~\mathrm{Cl^-(aq)}\\ \end{align*}\]
C
Net ionic equation
\[\begin{align*} \textit{no net ionic equation}\\ \end{align*}\]
D
\[\begin{align*} \textit{no precipitate} \end{align*}\]
E
\[\begin{align*} n(\mathrm{BaS}) &= n(\mathrm{BaCl_2}) ~ r(\mathrm{BaS,BaCl_2}) \\[1.5ex] &= m(\mathrm{BaCl_2}) ~ M(\mathrm{BaCl_2}) ~ r(\mathrm{BaS,BaCl_2}) \\[1.5ex] &= 5.00~\mathrm{g~BaCl_2} \left ( \dfrac{\mathrm{mol~BaCl_2}}{208.23~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{mol~BaS}}{\mathrm{mol~BaCl_2}} \right ) \\[1.5ex] &= 0.024\bar{0}11~\mathrm{mol}\\[3ex] n(\mathrm{BaS}) &= n(\mathrm{Na_2S}) ~ r(\mathrm{BaS,Na_2S}) \\[1.5ex] &= c(\mathrm{Na_2S}) ~ V(\mathrm{initial}) ~ r(\mathrm{BaS,Na_2S}) \\[1.5ex] &= \dfrac{1.40~\mathrm{mol~Na_2S}}{\mathrm{L}} \left ( \dfrac{225~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \left ( \dfrac{\mathrm{mol~BaS}}{\mathrm{mol~Na_2S}} \right ) \\[1.5ex] &= 0.31\bar{5}00~\mathrm{mol} \end{align*}\]
BaCl2 is limiting.
Concept: balancing equations; stoichiometry; molarity; limiting reactant; double-displacement reaction
A precipitate forms when titanium(IV) chloride is added to water. Two water molecules react and form four HCl molecules. What is the identity of the precipitate?
- What is the identity of the precipitate?
- What is the molar concentration (in mol L–1) of H+ ions if 2.00 g of TiCl4 was added to enough water to give a 100.0 mL solution?
Solution
Answer: TiCl4; 0.422 M
A
\[\begin{align*} \mathrm{TiCl_4(s)} + 2~\mathrm{H_2O(l)} &\longrightarrow 4~\mathrm{HCl(aq)} + \mathrm{TiO_2(s)} \\[1.5ex] \mathrm{TiO_2(s)} \end{align*}\]
B
\[\begin{align*} c(\mathrm{H^+}) &= n(\mathrm{H^+}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= n(\mathrm{HCl}) ~ r(\mathrm{H^+,HCl}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= n(\mathrm{TiCl_4}) ~ r(\mathrm{HCl,TiCl_4}) ~ r(\mathrm{H^+,HCl}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= m(\mathrm{TiCl_4}) ~ M(\mathrm{TiCl_4})^{-1} ~ r(\mathrm{HCl,TiCl_4}) ~ r(\mathrm{H^+,HCl}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= 2.00~\mathrm{g~TiCl_4} \left ( \dfrac{\mathrm{mol~TiCl_4}}{189.67~\mathrm{g}} \right ) \left ( \dfrac{4~\mathrm{mol~HCl}}{\mathrm{mol~TiCl_4}} \right ) \left ( \dfrac{\mathrm{mol~H^+}}{\mathrm{mol~HCl}} \right ) \biggr[ 100.0~\mathrm{mL} \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \biggr]^{-1} \\[1.5ex] &= 0.42\bar{1}78~\mathrm{mol~L^{-1}} = 0.422~\mathrm{mol~L^{-1}} \end{align*}\]
Concept: balancing equations; stoichiometry; molarity
A 123 mL sample of 0.210 M aqueous magnesium chloride forms a precipitate when mixed with 324 mL 0.120 M aqueous sodium hydroxide.
- How much (in g) precipitate is formed?
- What is the molar concentration (in mol L–1) of the magnesium(2+) ion?
- What is the molar concentration (in mol L–1) of the sodium(1+) ion?
Solution
Answer: 1.13 g; 0.0286 M; 0.0870 M
\[\begin{align*} \mathrm{MgCl_2(aq)} + 2~\mathrm{NaOH(aq)} &\longrightarrow 2~\mathrm{NaCl(aq)} + \mathrm{Mg(OH)_2(s)} \end{align*}\]
A
Find the limiting reactant
\[\begin{align*} n(\mathrm{Mg(OH)_2}) &= n(\mathrm{MgCl_2}) ~ r(\mathrm{Mg(OH)_2,MgCl_2})\\[1.5ex] &= c(\mathrm{MgCl_2}) ~ V(\mathrm{MgCl_2}) ~ r(\mathrm{Mg(OH)_2,MgCl_2}) \\[1.5ex] &= \dfrac{0.210~\mathrm{mol~MgCl_2}}{\mathrm{L}} \left ( \dfrac{123~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \left ( \dfrac{\mathrm{mol~Mg(OH)_2}}{\mathrm{mol~MgCl_2}} \right ) \\[1.5ex] &= 0.025\bar{8}3~\mathrm{mol} \\[3ex] n(\mathrm{Mg(OH)_2}) &= n(\mathrm{NaOH}) ~ r(\mathrm{Mg(OH)_2,NaOH})\\[1.5ex] &= c(\mathrm{NaOH}) ~ V(\mathrm{NaOH}) ~ r(\mathrm{Mg(OH)_2,NaOH}) \\[1.5ex] &= \dfrac{0.120~\mathrm{mol~NaOH}}{\mathrm{L}} \left ( \dfrac{324~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \left ( \dfrac{\mathrm{mol~Mg(OH)_2}}{2~\mathrm{mol~NaOH}} \right )\\[1.5ex] &= 0.019\bar{4}40~\mathrm{mol} \end{align*}\]
NaOH is limiting.
\[\begin{align*} m(\mathrm{Mg(OH)_2}) &= n(\mathrm{Mg(OH)_2}) ~ M(\mathrm{Mg(OH)_2})^{-1} \\[1.5ex] &= 0.019\bar{4}40~\mathrm{mol~Mg(OH)_2} \left ( \dfrac{58.33~\mathrm{g~Mg(OH)_2}}{\mathrm{mol}} \right )\\[1.5ex] &= 1.1\bar{3}39~\mathrm{g} = 1.13~\mathrm{g} \end{align*}\]
B
\[\begin{align*} c(\mathrm{Mg^{2+}}) &= n(\mathrm{Mg^{2+}})~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ n(\mathrm{Mg^{2+}})_{\mathrm{initial}} - n(\mathrm{Mg^{2+}})_{\mathrm{reacted}} \biggl\}~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ \biggr[ n(\mathrm{MgCl_2})_{\mathrm{initial}} ~ r(\mathrm{Mg^{2+},MgCl_2}) \biggr] - \biggr[ n(\mathrm{MgCl_2})_{\mathrm{final}}~ r(\mathrm{Mg^{2+},MgCl_2})~ \biggr] \biggl\} ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ \biggr[ n(\mathrm{Mg^{2+}})_{\mathrm{initial}}~ r(\mathrm{Cl^-,MgCl_2}) \biggr] - \biggr[ n(\mathrm{Mg(OH)_2})~ r(\mathrm{MgCl_2,Mg(OH)_2})~ r(\mathrm{Mg^{2+},MgCl_2}) \biggr] \biggl\} ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ \biggr[ c(\mathrm{MgCl_2})_{\mathrm{initial}}~ V(\mathrm{MgCl_2}) ~ r(\mathrm{Mg^{2+},MgCl_2}) \biggr] - \biggr[ n(\mathrm{Mg(OH)_2})~ r(\mathrm{MgCl_2,Mg(OH)_2})~ r(\mathrm{Mg^{2+},MgCl_2}) \biggr] \biggl\} \\ &\phantom{=}~~~~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ \biggr[ \dfrac{0.210~\mathrm{mol~MgCl_2}}{\mathrm{L}} \left ( \dfrac{123~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{\mathrm{10^3~mL}} \right ) \left ( \dfrac{\mathrm{mol~Mg^{2+}}}{\mathrm{mol~MgCl_2}} \right ) \biggr] - \\ &\phantom{=}~~~~~ \biggr[ 0.019\bar{4}40~\mathrm{mol~Mg(OH)_2} \left ( \dfrac{\mathrm{mol~MgCl_2}}{\mathrm{mol~Mg(OH)_2}} \right ) \left ( \dfrac{\mathrm{mol~Mg^{2+}}}{\mathrm{mol~MgCl_2}} \right ) \biggr] \biggl\} \left ( \dfrac{1}{(123 + 324)~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 0.014\bar{2}95~\mathrm{mol~L^{-1}} = 0.0143~\mathrm{mol~L^{-1}} \end{align*}\]
C
\[\begin{align*} c(\mathrm{Na^+}) &= n(\mathrm{Na^+}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= n(\mathrm{NaOH})_{\mathrm{initial}} ~ r(\mathrm{Na^+,NaOH}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= c(\mathrm{NaOH})_{\mathrm{initial}} ~ V(\mathrm{NaOH})_{\mathrm{initial}} ~ r(\mathrm{Na^+,NaOH}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \dfrac{0.120~\mathrm{mol~NaOH}}{\mathrm{L}} \left ( \dfrac{324~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{mol~Na^+}}{\mathrm{mol~NaOH}} \right ) \left ( \dfrac{1}{(123 + 324)~\mathrm{mL}} \right ) \\[1.5ex] &= 0.086\bar{9}79~\mathrm{mol~L^{-1}} = 0.0870~\mathrm{mol~L^{-1}} \end{align*}\]
Concept: balancing equations; stoichiometry; molarity
If 275 mL of a 0.125 M aqueous NaCl solution and 375 mL of a 0.575 M aqueous Na2SO4 solution are mixed, determine the molar concentrations (in mol L–1) of the following:
- chloride ions
- sulfate ions
- sodium ions
Solution
Answer: 0.0481 M; 0.332 M; 0.712 M
\[\begin{align*} \mathrm{NaCl(aq)} + \mathrm{Na_2SO_4(aq)} &\longrightarrow \mathrm{Na_2SO_4(aq)} + \mathrm{NaCl(aq)} \end{align*}\]
All compounds are soluble and there is no net ionic equation. Can determine ion molar concentrations without determining the limiting reactant.
A
\[\begin{align*} c(\mathrm{Cl^-}) &= n(\mathrm{Cl^-}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= c(\mathrm{NaCl}) ~ V(\mathrm{NaCl}) ~ r(\mathrm{Cl^-,NaCl}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \dfrac{0.125~\mathrm{mol~MgCl_2}}{\mathrm{L}} \left ( \dfrac{250.~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{mol~Cl^-}}{\mathrm{mol~NaOH}} \right ) \left ( \dfrac{1}{(275 + 375)~\mathrm{mL}} \right ) \\[1.5ex] &= 0.048\bar{0}76~\mathrm{mol~L^{-1}} = 0.0481~\mathrm{mol~L^{-1}} \end{align*}\]
B
\[\begin{align*} c(\mathrm{SO_4^{2-}}) &= n(\mathrm{SO_4^{2-}})~ V(\mathrm{solution})^{-1} \\[1.5ex] &= c(\mathrm{Na_2SO_4}) ~ V(\mathrm{Na_2SO_4}) ~ r(\mathrm{SO_4^{2-},Na_2SO_4}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \dfrac{0.575~\mathrm{mol~Na_2SO_4}}{\mathrm{L}} \left ( \dfrac{375~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{mol~SO_4^{2-}}}{\mathrm{mol~Na_2SO_4}} \right ) \left ( \dfrac{1}{(275 + 375)~\mathrm{mL}} \right ) \\[1.5ex] &= 0.33\bar{1}73~\mathrm{mol~L^{-1}} = 0.332~\mathrm{mol~L^{-1}} \end{align*}\]
C
\[\begin{align*} c(\mathrm{Na^+}) &= n(\mathrm{Na^+})~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ n(\mathrm{NaCl}) + n(\mathrm{Na_2SO_4}) \biggl\}~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ \biggr[ c(\mathrm{NaCl}) ~ V(\mathrm{NaCl}) ~ r(\mathrm{Na^+,NaCl}) ~ \biggr] + \biggr[ c(\mathrm{Na_2SO_4}) ~ V(\mathrm{Na_2SO_4}) ~ r(\mathrm{Na^+,Na_2SO_4}) ~ \biggr] \biggl\} \\ &\phantom{=}~~~~ V(\mathrm{solution})^{-1} \\[1.5ex] &= \biggl\{ \biggr[ \dfrac{0.125~\mathrm{mol~NaCl}}{\mathrm{L}} \left ( \dfrac{250.~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{mol~Na^+}}{\mathrm{mol~NaCl}} \right ) \biggr] + \\ &\phantom{=}~~~~ \biggr[ \dfrac{0.575~\mathrm{mol~Na_2SO_4}}{\mathrm{L}} \left ( \dfrac{375~\mathrm{mL}}{} \right ) \left ( \dfrac{2~\mathrm{mol~Na^+}}{\mathrm{mol~Na_2SO_4}} \right ) \biggr] \biggl\} \\ &\phantom{=}~~~~ \left ( \dfrac{1}{(275 + 375)~\mathrm{mL}} \right ) \\[1.5ex] &= 0.71\bar{1}53~\mathrm{mol~L^{-1}} = 0.712~\mathrm{mol~L^{-1}} \end{align*}\]
Concept: balancing equations; stoichiometry; molarity
Thermochemistry Basics
Sugar is melted in a pot and its temperature is measured as it heats. In this scenario, what is the system?
- the pot and sugar
- the stove
- the entire kitchen
- the rest of the universe
Solution
Answer: A
Concept: thermochemistry basics
Which of the following is not a type of potential energy (select all that apply)?
- Energy held in chemical bonds
- Energy resulting from intramolecular attractions
- Energy from the random motion of molecules
- Energy of a ball dropping from a height
Solution
Answer: C, D
Concept: thermochemistry basics
Which of the following is true of heat (select all that apply)?
- Heat is a form of thermal energy.
- Heat is the transfer of thermal energy.
- Heat is the action of forces through a distance.
- A negative heat in the system means the surroundings loses energy.
- A negative heat in the system means the system loses energy.
Solution
Answer: B, E
Concept: thermochemistry basics
Which of the following are false (select all that apply)?
- Energy can be converted from one type to another.
- Energy is the capacity to do work.
- Kinetic energy is energy resulting from condition, position, or composition.
- Potential energy is energy transferred between a system and its surroundings as a result of a temperature difference.
Solution
Answer: C, D
Concept: thermochemistry basics
A block of ice absorbs heat and melts. The value q for the system is:
- Positive
- Negative
- Zero
- There is not enough information to determine.
Solution
Answer: A
Concept: thermochemistry basics
Questions are written by UGA Chemistry unless otherwise indicated and translated with minor tweaks into an online format by Eric Van Dornshuld.