Stoichiometry

Chapter 04

Audio Summary

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Stoichiometry deals with the relationship between the amounts of reactants and products in a chemical equation.

A Recipe

A chemical equation can be thought of as a “recipe” where the ingredients are the ratio of reactants with which to make products.

Sometimes a recipe calls for a distinct amount of something. For example, a cookie dough recipe may require “2 eggs”. For ingredients that cannot be counted, volume or mass is listed such as “2 cups of flour” or “20 g of cornstarch”.

Chemical equations always give the exact number of particles required, whether interpreted as individual particles or moles of particles.

\[2~\mathrm{H_2(g)} + \mathrm{O_2(g)} \longrightarrow 2~\mathrm{H_2O(l)}\]

Here, two particles of dihydrogen and one particle of dioxygen is required to make two particles of water. Alternatively, two moles of dihydrogen and one moles of dioxygen is required to make two moles of water.

Working with Moles

If we wanted to make 2 moles of water using the reaction above, how much (in mol) of dihydrogen and oxygen would we need? We can use the stoichiometric factor (i.e. mole ratio, rB,A) of reactants to products to determine this.

\[\begin{align*} n(\mathrm{H_2}) &= n(\mathrm{H_2O}) ~ r(\mathrm{H_2, H_2O}) \\[1.5ex] &= \left (2~\mathrm{mol~H_2O} \right ) \left ( \dfrac{2~\mathrm{mol~H_2}}{2~\mathrm{mol~H_2O}} \right ) \\[1.5ex] &= 2~\mathrm{mol} \\[3ex] n(\mathrm{O_2}) &= n(\mathrm{H_2O}) ~ r(\mathrm{O_2, H_2O}) \\[1.5ex] &= \left ( 2~\mathrm{mol~H_2O} \right ) \left ( \dfrac{1~\mathrm{mol~O_2}}{2~\mathrm{mol~H_2O}} \right ) \\[1.5ex] &= 1~\mathrm{mol} \end{align*}\]

Let’s write a[n] (IRF) table tracking the moles of reactants and products for our reaction.

Initial (mol) 2
1
0
Reaction 2H2 + O2 2H2O
Final (mol) 0
0
2

Working with Mass

When performing a controlled reaction, one does not “count” the particles with which to mix. Rather, we weigh or measure (by volume or pressure) the amounts of reactants.

Determine required amount of reactants to make product

If we wanted to make 2 moles of water using the reaction above, what mass (in g) of dihydrogen and oxygen would we need? We can use the molar mass of each substance and stoichiometric factor (i.e. mole-to-mole ratio) of reactants to products to determine this.

\[\begin{align*} m(\mathrm{H_2}) &= n(\mathrm{H_2O}) ~ r(\mathrm{H_2, H_2O}) ~ M(\mathrm{H_2}) \\[1.5ex] &= \left (2~\mathrm{mol~H_2O} \right ) \left ( \dfrac{2~\mathrm{mol~H_2}}{2~\mathrm{mol~H_2O}} \right ) \left ( \dfrac{2.02~\mathrm{g}}{\mathrm{mol~H_2}} \right ) \\[1.5ex] &= 4.04~\mathrm{g} \\[3ex] m(\mathrm{O_2}) &= n(\mathrm{H_2O}) ~ r(\mathrm{O_2, H_2O}) ~ M(\mathrm{O_2}) \\[1.5ex] &= \left (2~\mathrm{mol~H_2O} \right ) \left ( \dfrac{1~\mathrm{mol~O_2}}{2~\mathrm{mol~H_2O}} \right ) \left ( \dfrac{32.00~\mathrm{g}}{\mathrm{mol~O_2}} \right ) \\[1.5ex] &= 32.00~\mathrm{g} \end{align*}\]


What mass (in g) of water was produced?

\[\begin{align*} m(\mathrm{H_2O}) &= n(\mathrm{H_2O}) ~ M(\mathrm{H_2O}) \\[1.5ex] &= (2~\mathrm{mol~H_2O}) \left ( \dfrac{18.02~\mathrm{g}}{\mathrm{mol}} \right ) \\[1.5ex] &= 36.04~\mathrm{g} \end{align*}\]


We now know the masses of each ingredient of this recipe to make 36.04 g of water. We could write our recipe with their corresponding masses such as below.

\[4.04~\mathrm{g}~\mathrm{H_2(g)} + 32.00~\mathrm{g}~\mathrm{O_2(g)} \longrightarrow 36.04~\mathrm{g}~\mathrm{H_2O(l)}\]

However, this notation is rather inconvenient. We know that microscopically, substances are discrete particles and reactions between particles depend on the number of particles that are present.

Balanced reactions accurately show the ratio of these particles. In other words, if making a sandwich, it is more important to know how many of each ingredient you need (discrete number) rather than their masses (e.g. its more valuable having two slices of bread rather than 200 g of bread). If we knew each slice of bread was 100 g (analogous to molar mass), then we would know that 200 g of bread could net us a sandwich. We use molar masses to quickly convert between moles of an ingredient to the mass that would need to be weighed.


Reactant masses to make exactly 1 mole of water?

\[\begin{align*} m(\mathrm{H_2}) &= n(\mathrm{H_2O}) ~ r(\mathrm{H_2, H_2O}) ~ M(\mathrm{H_2}) \\[1.5ex] &= \left (1~\mathrm{mol~H_2O} \right ) \left ( \dfrac{2~\mathrm{mol~H_2}}{2~\mathrm{mol~H_2O}} \right ) \left ( \dfrac{2.02~\mathrm{g}}{\mathrm{mol~H_2}} \right ) \\[1.5ex] &= 2.02~\mathrm{g} \\[3ex] m(\mathrm{O_2}) &= n(\mathrm{H_2O}) ~ r(\mathrm{O_2, H_2O}) ~ M(\mathrm{O_2}) \\[1.5ex] &= \left (1~\mathrm{mol~H_2O} \right ) \left ( \dfrac{1~\mathrm{mol~O_2}}{2~\mathrm{mol~H_2O}} \right ) \left ( \dfrac{32.00~\mathrm{g}}{\mathrm{mol~O_2}} \right ) \\[1.5ex] &= 16.00~\mathrm{g} \end{align*}\]

Reactant masses to make exactly 15 moles of water?

\[\begin{align*} m(\mathrm{H_2}) &= n(\mathrm{H_2O}) ~ r(\mathrm{H_2, H_2O}) ~ M(\mathrm{H_2}) \\[1.5ex] &= \left ( 15~\mathrm{mol~H_2O} \right ) \left ( \dfrac{2~\mathrm{mol~H_2}}{2~\mathrm{mol~H_2O}} \right ) \left ( \dfrac{2.02~\mathrm{g}}{\mathrm{mol~H_2}} \right ) \\[1.5ex] &= 30.3~\mathrm{g} \\[3ex] m(\mathrm{O_2}) &= n(\mathrm{H_2O}) ~ r(\mathrm{O_2, H_2O}) ~ M(\mathrm{O_2}) \\[1.5ex] &= \left ( 15~\mathrm{mol~H_2O} \right ) \left ( \dfrac{1~\mathrm{mol~O_2}}{2~\mathrm{mol~H_2O}} \right ) \left ( \dfrac{32.00~\mathrm{g}}{\mathrm{mol~O_2}} \right ) \\[1.5ex] &= 240.0~\mathrm{g} \end{align*}\]

Reactant masses to make exactly 3.54 moles of water?

\[\begin{align*} m(\mathrm{H_2}) &= n(\mathrm{H_2O}) ~ r(\mathrm{H_2, H_2O}) ~ M(\mathrm{H_2}) \\[1.5ex] &= \left ( 3.54~\mathrm{mol~H_2O} \right ) \left ( \dfrac{2~\mathrm{mol~H_2}}{2~\mathrm{mol~H_2O}} \right ) \left ( \dfrac{2.02~\mathrm{g}}{\mathrm{mol~H_2}} \right ) \\[1.5ex] &= 7.1\bar{5}0~\mathrm{g} \\[3ex] m(\mathrm{O_2}) &= n(\mathrm{H_2O}) ~ r(\mathrm{O_2, H_2O}) ~ M(\mathrm{O_2}) \\[1.5ex] &= \left ( 3.54~\mathrm{mol~H_2O} \right ) \left ( \dfrac{1~\mathrm{mol~O_2}}{2~\mathrm{mol~H_2O}} \right ) \left ( \dfrac{32.00~\mathrm{g}}{\mathrm{mol~O_2}} \right ) \\[1.5ex] &= 56.\bar{6}4~\mathrm{g} \end{align*}\]


Determine amount of product that can be made from a limited amount of reactant(s)

So far we started with the chemical equation (our recipe) and determined the required mass of each reactant to make a certain amount of water.

What if instead we went to our chemical cabinet and found that we had a limited amount of dihydrogen on hand and an abundance of dioxygen? We could figure out how much water we could make based on how much reactant we have.

How much water (in g) could be made if starting with 512.5 g of dihydrogen and an abundance of dioxygen? In this example, dihydrogen is the limiting reactant, the reactant present in the least amount that dictates the amount of product that could be made.

\[\begin{align*} n(\mathrm{H_2O}) &= m(\mathrm{H_2}) ~ M(\mathrm{H_2})^{-1} ~ r(\mathrm{H_2O, H_2}) ~ \\[1.5ex] &= \left ( 512.5~\mathrm{g~H_2} \right ) \left ( \dfrac{\mathrm{mol~H_2}}{2.02~\mathrm{g}} \right ) \left ( \dfrac{2~\mathrm{mol~H_2O}}{2~\mathrm{mol~H_2}} \right ) \\[1.5ex] &= 25\bar{3}.71~\mathrm{mol~H_2O} \\[3ex] m(\mathrm{H_2O}) &= n(\mathrm{H_2O}) ~ M(\mathrm{H_2O}) \\[1.5ex] &= \left ( 25\bar{3}.7~\mathrm{mol~H_2O} \right ) \left ( \dfrac{18.02~\mathrm{g}}{\mathrm{mol~H_2O}} \right ) \\[1.5ex] &= 4~5\bar{7}1.8~\mathrm{g} \\[1.5ex] &= 4.57\times 10^{3}~\mathrm{g} \end{align*}\]


Lets combine both equations above into one equation to see the dimensional analysis in one complete expression.

\[\begin{align*} m(\mathrm{H_2O}) &= m(\mathrm{H_2}) ~ M(\mathrm{H_2})^{-1} ~ r(\mathrm{H_2O, H_2}) ~ M(\mathrm{H_2O}) \\[1.5ex] &= \left ( 512.5~\mathrm{g~H_2} \right ) \left ( \dfrac{\mathrm{mol~H_2}}{2.02~\mathrm{g}} \right ) \left ( \dfrac{2~\mathrm{mol~H_2O}}{2~\mathrm{mol~H_2}} \right ) \left ( \dfrac{18.02~\mathrm{g}}{\mathrm{mol~H_2O}} \right ) \\[1.5ex] &= 4~5\bar{7}1.8~\mathrm{g} \\[1.5ex] &= 4.57\times 10^{3}~\mathrm{g} \end{align*}\]

The operations that took place are as follows:

  1. The mass of H2 is converted to moles by dividing by the molar mass of H2
  2. The moles of H2 was converted to the number of moles of H2O by using the stoichiometric factor for H2 and H2O as given by the balanced chemical equation
  3. The moles of H2O was converted to mass of H2O by multiplying by the molar mass of H2O

Stoichiometry Relationships


Flowchart of stiochiometric relationships and other quantities

Flowchart of stiochiometric relationships and other quantities



Reaction Data

The following examples illustrate how to use the stoichiometric flowchart.

A sample of aqueous nitric acid is mixed with enough aqueous magnesium hydroxide [d(Mg(OH)2) = 2.3446 g cm–3] to undergo a complete reaction (at 25 °C) to produce magnesium nitrate [d(Mg(NO3)2) = 2.3 g cm–3]. The final volume of the solution is 325 mL.

The aqueous nitric acid sample was prepared by mixing 15.00 mL of anhydrous HNO3 [d(HNO3) = 1.5129 g mL–1] with enough water [d(H2O) = 0.99713 g mL–1] to make a 0.250 L solution.

From the statement above, you should be able to sketch out a balanced chemical equation for the reaction.

\[2~\mathrm{HNO_3(aq)} + \mathrm{Mg(OH)_2(aq)} \longrightarrow 2~\mathrm{H_2O(l)} + \mathrm{Mg(NO_3)_2(aq)}\]


We will now explore different questions that could be asked followed by how to determine the answers. We will calculate mass (m), moles (n), volume (V), and number of particles (N) for each reactant and product.


Volume to mass: V(HNO3) → m(HNO3)

What is the mass (in g) of nitric acid?

To convert volume of nitric acid to mass, use the density of nitric acid which is given as 1.5129 g mL–1.

\[\begin{align*} d(\mathrm{HNO_3}) &= \dfrac{m(\mathrm{HNO_3})}{V(\mathrm{HNO_3})} \longrightarrow \\[1.5ex] m(\mathrm{HNO_3}) &= d(\mathrm{HNO_3}) ~ V(\mathrm{HNO_3}) \\[1.5ex] &= \dfrac{1.5129~\mathrm{g}}{\mathrm{mL}} \left ( 15.00~\mathrm{mL}\right ) \\[1.5ex] &= 22.6\bar{9}4~\mathrm{g} \end{align*}\]


Mass to moles: m(HNO3) → n(HNO3)

How many moles of nitric acid is present?

Convert the mass of nitric acid to moles by dividing by the molar mass of nitric acid.

\[\begin{align*} n(\mathrm{HNO_3}) &= m(\mathrm{HNO_3}) ~ M(\mathrm{HNO_3})^{-1} \\[1.5ex] &= 22.6\bar{9}4~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{63.01~\mathrm{g}} \right ) \\[1.5ex] &= 0.360\bar{1}6~\mathrm{mol} \end{align*}\]


Moles to number of particles: n(HNO3) → N(HNO3)

How many particles of nitric acid is consumed by the reaction?

Multiply the moles of nitric acid by Avogadro’s number.

\[\begin{align*} N(\mathrm{HNO_3}) &= n(\mathrm{HNO_3}) ~ N_{\mathrm{A}} \\[1.5ex] &= 0.360\bar{1}6~\mathrm{mol} \left ( \dfrac{6.022\times 10^{23}}{\mathrm{mol}} \right ) \\[1.5ex] &= 2.16\bar{8}8\times 10^{23} \end{align*}\]


Moles to moles: n(HNO3) → n(Mg(OH)2)

How many moles of magnesium hydroxide is needed to react with all of the nitric acid?

Use the stiochiometric factor relating the moles of nitric acid and magnesium hydroxide.

\[\begin{align*} n(\mathrm{Mg(OH)_2}) &= n(\mathrm{HNO_3}) ~ r(\mathrm{Mg(OH)_2, HNO_3}) \\[1.5ex] &= 0.360\bar{1}6~\mathrm{mol} \left ( \dfrac{\mathrm{1~mol~Mg(OH)_2}}{2~\mathrm{mol~HNO_3}}\right ) \\[1.5ex] &= 0.180\bar{0}8~\mathrm{mol} \end{align*}\]


Moles to mass: n(Mg(OH)2) → m(Mg(OH)2)

What mass (in g) of Mg(OH)2 is needed to completely react with the nitric acid?

Multiply the number of moles of Mg(OH)2 by the molar mass.

\[\begin{align*} m(\mathrm{Mg(OH)_2}) &= n(\mathrm{Mg(OH)_2}) ~ M(\mathrm{Mg(OH)_2}) \\[1.5ex] &= 0.180\bar{0}8~\mathrm{mol} \left ( \dfrac{\mathrm{58.32~g}}{\mathrm{mol~Mg(OH)_2}} \right ) \\[1.5ex] &= 10.5\bar{0}2~\mathrm{g} \end{align*}\]


Mass to volume: m(Mg(OH)2) → V(Mg(OH)2)

What volume (in cm3) of magnesium hydroxide is needed to completely react with nitric acid?

Use the density of magnesium hydroxide to find the volume.

\[\begin{align*} d(\mathrm{Mg(OH)_2}) &= \dfrac{m(\mathrm{Mg(OH)_2})}{V(\mathrm{Mg(OH)_2})} \longrightarrow \\[1.5ex] V(\mathrm{Mg(OH)_2}) &= \dfrac{m(\mathrm{Mg(OH)_2})}{d(\mathrm{Mg(OH)_2})} \\[1.5ex] &= \dfrac{10.5\bar{0}2~\mathrm{g}}{2.3446~\mathrm{g~cm^{-3}}} \\[1.5ex] &= 4.47\bar{9}2~\mathrm{cm^3} \end{align*}\]


Moles to number of particles: n(Mg(OH)2) → N(Mg(OH)2)

How many particles of magnesium hydroxide is consumed by the reaction?

Multiply the moles of magnesium hydroxide by Avogadro’s number.

\[\begin{align*} N(\mathrm{Mg(OH)_2}) &= n(\mathrm{Mg(OH)_2}) ~ N_{\mathrm{A}} \\[1.5ex] &= 0.180\bar{0}8~\mathrm{mol} \left ( \dfrac{6.022\times 10^{23}}{\mathrm{mol}} \right ) \\[1.5ex] &= 1.08\bar{4}4\times 10^{23} \end{align*}\]


Moles to moles: n(HNO3) → n(H2O)

How many moles of water is produced from the reaction?

Use the stiochiometric factor relating the moles of nitric acid and water.

\[\begin{align*} n(\mathrm{H_2O}) &= n(\mathrm{HNO_3}) ~ r(\mathrm{H_2O, HNO_3}) \\[1.5ex] &= 0.360\bar{1}6~\mathrm{mol} \left ( \dfrac{2~\mathrm{mol~H_2O}}{2~\mathrm{mol~HNO_3}}\right ) \\[1.5ex] &= 0.360\bar{1}6~\mathrm{mol} \end{align*}\]


Moles to mass: n(H2O) → m(H2O)

What mass of water (in g) is produced by the reaction?

Convert moles of water to mass by multiplying by the molar mass of water.

\[\begin{align*} m(\mathrm{H_2O}) &= n(\mathrm{H_2O}) ~ M(\mathrm{H_2O}) \\[1.5ex] &= 0.360\bar{1}6~\mathrm{mol} \left ( \dfrac{\mathrm{18.02~g}}{\mathrm{mol~H_2O}} \right ) \\[1.5ex] &= 6.49\bar{0}0~\mathrm{g} \end{align*}\]


Mass to volume: m(H2O) → V(H2O)

What volume (in mL) of water is produced by the reaction?

Use the density of water to find the volume.

\[\begin{align*} d(\mathrm{H_2O}) &= \dfrac{m(\mathrm{H_2O})}{V(\mathrm{H_2O})} \longrightarrow \\[1.5ex] V(\mathrm{H_2O}) &= \dfrac{m(\mathrm{H_2O})}{d(\mathrm{H_2O})} \\[1.5ex] &= \dfrac{6.49\bar{0}0~\mathrm{g}}{0.99713~\mathrm{g~mL^{-1}}} \\[1.5ex] &= 6.50\bar{8}7~\mathrm{mL} \end{align*}\]


Moles to number of particles: n(H2O) → N(H2O)

How many particles of water is produced by the reaction?

Multiply the moles of water by Avogadro’s number.

\[\begin{align*} N(\mathrm{H_2O}) &= n(\mathrm{H_2O}) ~ N_{\mathrm{A}} \\[1.5ex] &= 0.360\bar{1}6~\mathrm{mol} \left ( \dfrac{6.022\times 10^{23}}{\mathrm{mol}} \right ) \\[1.5ex] &= 2.16\bar{8}8\times 10^{23} \end{align*}\]


Moles to moles: n(HNO3) → n(Mg(NO3)2)

How many moles of magnesium nitrate is produced from the reaction?

Convert moles of nitric acid to moles of magnesium nitrate by using the stoichiometric factor.

\[\begin{align*} n(\mathrm{Mg(NO_3)_2}) &= n(\mathrm{HNO_3}) ~ r(\mathrm{Mg(NO_3)_2, HNO_3}) \\[1.5ex] &= 0.360\bar{1}6~\mathrm{mol} \left ( \dfrac{1~\mathrm{mol~Mg(NO_3)_2}}{2~\mathrm{mol~HNO_3}}\right ) \\[1.5ex] &= 0.180\bar{0}8~\mathrm{mol} \end{align*}\]


Moles to mass: n(Mg(NO3)2) → m(Mg(NO3)2)

What mass of magnesium nitrate (in g) is produced by the reaction?

Convert moles of magnesium nitrate to mass by multiplying by the molar mass of magnesium nitrate.

\[\begin{align*} m(\mathrm{Mg(NO_3)_2}) &= n(\mathrm{Mg(NO_3)_2}) ~ M(\mathrm{Mg(NO_3)_2}) \\[1.5ex] &= 0.180\bar{0}8~\mathrm{mol} \left ( \dfrac{\mathrm{148.32~g}}{\mathrm{mol~Mg(NO_3)_2}} \right ) \\[1.5ex] &= 26.7\bar{0}9~\mathrm{g} \end{align*}\]


Mass to volume: m(Mg(NO3)2) → V(Mg(NO3)2)

What volume (in cm3) of magnesium nitrate is produced by the reaction?

Use the density of magnesium nitrate to find the volume.

\[\begin{align*} d(\mathrm{Mg(NO_3)_2}) &= \dfrac{m(\mathrm{Mg(NO_3)_2})}{V(\mathrm{Mg(NO_3)_2})} \longrightarrow \\[1.5ex] V(\mathrm{Mg(NO_3)_2}) &= \dfrac{m(\mathrm{Mg(NO_3)_2})}{d(\mathrm{Mg(NO_3)_2})} \\[1.5ex] &= \dfrac{26.7\bar{0}9~\mathrm{g}}{2.3~\mathrm{g~cm^{-3}}} \\[1.5ex] &= 1\bar{1}.6~\mathrm{cm^3} \end{align*}\]


Moles to number of particles: n(Mg(NO3)2) → N(Mg(NO3)2)

How many particles of magnesium nitrate is produced by the reaction?

Multiply the moles of magnesium nitrate by Avogadro’s number.

\[\begin{align*} N(\mathrm{Mg(NO_3)_2}) &= n(\mathrm{Mg(NO_3)_2}) ~ N_{\mathrm{A}} \\[1.5ex] &= 0.180\bar{0}8~\mathrm{mol} \left ( \dfrac{6.022\times 10^{23}}{\mathrm{mol}} \right ) \\[1.5ex] &= 1.08\bar{4}4\times 10^{23} \end{align*}\]


Moles to molar concentration: n(HNO3) → c(HNO3)

What is the molar concentration (c in mol L–1) of the nitric acid solution?

See molar concentration.

Divide the number of moles of nitric acid by the volume (in L) of the nitric acid solution.

\[\begin{align*} c(\mathrm{HNO_3}) &= \dfrac{n(\mathrm{HNO_3})}{V_{\mathrm{soln}}} \\[1.5ex] &= \dfrac{0.360\bar{1}6~\mathrm{mol}}{0.250~\mathrm{L}} \\[1.5ex] &= 1.4\bar{4}0~\mathrm{mol~L^{-1}} \end{align*}\]


Summary of Data

Table 1: Data for reaction
Property HNO3 + Mg(OH)2 H2O + Mg(NO3)2
m (g) 22.69
10.50
6.490
26.71
V (mL or cm3) 15.00
4.479
6.509
11.6
n (mol) 0.3602
0.1801
0.3602
0.1801
N 2.169 × 1023
1.084 × 1023
2.169 × 1023
1.084 × 1023

We could write an IRF table showing the initial and final number of moles of reactants and products for this reaction. Note: This IRF table does not indicate the excess magnesium hydroxide in solution, only the number of moles of sodium hydroxide that was consumed.

Initial (mol) 0.3602
0.1801
0
0
Reaction 2 HNO3(aq) + Mg(OH)2(aq) 2 H2O(l) + Mg(NO3)2(aq)
Final (mol) 0
0
0.3602
0.1801


IRF Tables (in other quantities)

We could write an IRF table that shows the initial and final mass (m), volume (V), or number of particles N, for the reactants and products, though this is not particularly useful.

Initial (g) 22.69
10.50
0
0
Reaction 2 HNO3(aq) + Mg(OH)2(aq) 2 H2O(l) + Mg(NO3)2(aq)
Final (g) 0
0
6.490
26.71


Initial (mL or cm3) 15.00
4.479
0
0
Reaction 2 HNO3(aq) + Mg(OH)2(aq) 2 H2O(l) + Mg(NO3)2(aq)
Final (mL or cm3) 0
0
6.509
11.6


Initial (N) 2.169 × 1023
1.084 × 1023
0
0
Reaction 2 HNO3(aq) + Mg(OH)2(aq) 2 H2O(l) + Mg(NO3)2(aq)
Final (N) 0
0
2.169 × 1023
1.084 × 1023

Additional Analyses

Let us ask some other questions regarding our reaction.

What is the limiting reactant?

According to the question, a sample of nitric acid is mixed with “enough” magnesium hydroxide to undergo a complete reaction. Based on the wording, nitric acid is the limiting reactant.


How much magnesium hydroxide is leftover?

Let’s specify the amount of magnesium hydroxide that is initially present as 0.5000 mol.

If there was 0.5000 mol of Mg(OH)2 present, how many moles of magnesium hydroxide would be leftover from the reaction?

Determine the amount of Mg(OH)2 leftover (i.e. final amount) from the reaction.

\[\begin{align*} n_{\mathrm{final}}(\mathrm{Mg(OH)_2}) &= n_{\mathrm{initial}}(\mathrm{Mg(OH)_2}) - n_{\mathrm{consumed}}(\mathrm{Mg(OH)_2}) \\[1.5ex] &= 0.5000~\mathrm{mol} - 0.180\bar{0}8~\mathrm{mol} \\[1.5ex] &= 0.319\bar{9}2~\mathrm{mol} \end{align*}\]

Let’s write another IRF table that illustrates the actual amounts (in mol) of magnesium hydroxide before and after the reaction.

Initial (mol) 0.3602
0.5000
0
0
Reaction 2 HNO3(aq) + Mg(OH)2(aq) 2 H2O(l) + Mg(NO3)2(aq)
Final (mol) 0
0.3199
0.3602
0.1801


How much hydroxide is leftover?

Determine the moles of hydroxide leftover by using the stoichiometric factor of hydroxide to magnesium hydroxide.

\[\begin{align*} n(\mathrm{OH^-}) &= n_{\mathrm{final}}(\mathrm{Mg(OH)_2}) ~ r(\mathrm{OH^-, Mg(OH)_2}) \\[1.5ex] &= 0.319\bar{9}2~\mathrm{mol} \left ( \dfrac{2~\mathrm{mol~OH^-}}{\mathrm{1~mol~Mg(OH)_2}} \right ) \\[1.5ex] &= 0.639\bar{8}4~\mathrm{mol} \end{align*}\]


What is the final molar concentration of hydroxide?

Divide the moles of leftover hydroxide by the total volume of solution (in L).

\[\begin{align*} c(\mathrm{OH^-}) &= \dfrac{n(\mathrm{OH^-})}{V} \\[1.5ex] &= \dfrac{0.639\bar{8}4~\mathrm{mol~OH^-}}{0.325~\mathrm{L}} \\[1.5ex] &= 1.9\bar{6}8~\mathrm{mol~L^{-1}} \\[1.5ex] &= 1.97~M \end{align*}\]

What are the initial molar concentrations of the solutes?

To find the molar concentration (mol L–1) of each solute in the solution, divide the initial number of moles of each species by the total volume of solution (in L). The problem states that the final reaction mixture is 325 mL or 0.325 L.

\[\begin{align*} c(\mathrm{HNO_3}) &= \dfrac{n(\mathrm{HNO_3})}{V_{\mathrm{soln}}} \\[1.5ex] &= \dfrac{0.360\bar{1}6~\mathrm{mol}}{0.325~\mathrm{L}} \\[1.5ex] &= 1.1\bar{0}8~\mathrm{mol~L^{-1}} \\[1.5ex] &= 1.1\bar{0}8~M\\[3ex] c(\mathrm{Mg(OH)_2}) &= \dfrac{n(\mathrm{Mg(OH)_2})}{V_{\mathrm{soln}}} \\[1.5ex] &= \dfrac{0.5000~\mathrm{mol}}{0.325~\mathrm{L}} \\[1.5ex] &= 1.5\bar{3}8~\mathrm{mol~L^{-1}}\\[1.5ex] &= 1.5\bar{3}8~M \end{align*}\]


What are the final molar concentrations of the remaining solutes?

To find the molar concentration (mol L–1) of each solute in the solution, divide the final number of moles of each species by the total volume of solution (in L). The problem states that the final reaction mixture is 325 mL or 0.325 L.

\[\begin{align*} c(\mathrm{Mg(OH)_2}) &= \dfrac{n(\mathrm{Mg(OH)_2})}{V_{\mathrm{soln}}} \\[1.5ex] &= \dfrac{0.319\bar{9}2~\mathrm{mol}}{0.325~\mathrm{L}} \\[1.5ex] &= 0.98\bar{4}3~\mathrm{mol~L^{-1}}\\[1.5ex] &= 0.98\bar{4}3~M\\[3ex] c(\mathrm{Mg(NO_3)_2}) &= \dfrac{n(\mathrm{Mg(NO_3)_2}}{V_{\mathrm{soln}}} \\[1.5ex] &= \dfrac{0.180\bar{0}8~\mathrm{mol}}{0.325~\mathrm{L}} \\[1.5ex] &= 0.55\bar{4}0~\mathrm{mol~L^{-1}}\\[1.5ex] &= 0.55\bar{4}0~M \end{align*}\]

We can write an IRF table to show the initial and final molar concentrations of the reactants and products.

Initial (M) 1.11
1.54


0
Reaction 2 HNO3(aq) + Mg(OH)2(aq) 2 H2O(l) + Mg(NO3)2(aq)
Final (M) 0
0.984


0.554


Percent Yield

What is the percent yield of the reaction if 25.11 g Mg(NO3)2 was obtained?

If 22.69 g of nitric acid reacts with 10.50 g of magnesium hydroxide, 6.490 g of water and 26.71 g of magnesium nitrate should be produced. Let’s say you run the reaction and only collect 25.11 g of magnesium nitrate. What is the percent yield of the reaction?

\[\begin{align*} \%~\mathrm{yield} &= \dfrac{\mathrm{actual~yield}}{\mathrm{theoretical~yield}} \times 100\% \\[1.5ex] &= \dfrac{\mathrm{25.11~\mathrm{g}}}{\mathrm{26.71~\mathrm{g}}} \times 100\% \\[1.5ex] &= 94.0\bar{0}9~\% \\[1.5ex] &= 94.01~\% \end{align*}\]


Practice


Continuing with the reaction introduced above, how much hydroxide (in g) is leftover if the percent yield was only 89.5 %?

Solution

Determine the actual yield (in g) of one of the products. Here I choose Mg(NO3)2 as my product.

\[\begin{align*} \%~\mathrm{yield} &= \dfrac{\mathrm{actual~yield}}{\mathrm{theoretical~yield}} \times 100\% \longrightarrow \\[2.5ex] \mathrm{actual~yield~Mg(NO_3)_2} &= \left ( \dfrac{\%~\mathrm{yield}}{100~\%} \right ) \left( \mathrm{theoretical~yield~Mg(NO_3)_2} \right ) \\[1.5ex] &= \left ( \dfrac{89.5~\%}{100~\%} \right ) \mathrm{26.71~\mathrm{g}} \\[1.5ex] &= 23.\bar{9}0~\mathrm{g} \end{align*}\]

Next, find the number of moles of magnesium nitrate that was actually produced.

\[\begin{align*} n(\mathrm{Mg(NO_3)_2}) &= m(\mathrm{Mg(NO_3)_2}) ~ M(\mathrm{Mg(NO_3)_2})^{-1} \\[1.5ex] &= 23.\bar{9}0~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{148.32~\mathrm{g}} \right ) \\[1.5ex] &= 0.16\bar{1}1~\mathrm{mol} \end{align*}\]

Now, find the moles of magnesium hydroxide that was actually consumed by using the stoichiometric factor.

\[\begin{align*} n(\mathrm{Mg(OH)_2}) &= n(\mathrm{Mg(NO_3)_2}) ~ r(\mathrm{Mg(OH)_2, Mg(NO_3)_2}) \\[1.5ex] &= 0.16\bar{1}1~\mathrm{mol} \left ( \dfrac{1~\mathrm{mol~Mg(OH)_2}}{1~\mathrm{mol~Mg(NO_3)_2}} \right ) \\[1.5ex] &= 0.16\bar{1}1~\mathrm{mol} \end{align*}\]

Determine the amount of magnesium hydroxide that is leftover from the reaction. Recall that there was 0.5000 mol of Mg(OH)2 present.

\[\begin{align*} n_{\mathrm{leftover}}(\mathrm{Mg(OH)_2}) &= n_{\mathrm{available}}(\mathrm{Mg(OH)_2}) - n_{\mathrm{consumed}}(\mathrm{Mg(OH)_2}) \\[1.5ex] &= 0.5000~\mathrm{mol} - 0.16\bar{1}1~\mathrm{mol} \\[1.5ex] &= 0.33\bar{8}9~\mathrm{mol} \end{align*}\]

Convert the moles of magnesium hydroxide to moles of hydroxide.

\[\begin{align*} n(\mathrm{OH^-}) &= n(\mathrm{Mg(OH)_2}) \left ( \dfrac{2~\mathrm{mol~OH^-}}{1~\mathrm{mol~Mg(OH)_2}} \right ) \\[1.5ex] &= 0.33\bar{8}9~\mathrm{mol} \left ( \dfrac{2~\mathrm{mol~OH^-}}{1~\mathrm{mol~Mg(OH)_2}} \right ) \\[1.5ex] &= 0.67\bar{7}8~\mathrm{mol} \end{align*}\]

Next, find the mass (in g) of the remaining hydroxide by multiplying by the molar mass.

\[\begin{align*} m(\mathrm{OH^-}) &= n(\mathrm{OH^-}) ~ M(\mathrm{OH^-}) \\[1.5ex] &= 0.67\bar{7}8~\mathrm{mol} \left ( \dfrac{17.01~\mathrm{g}}{\mathrm{mol}} \right ) \\[1.5ex] &= 11.\bar{5}2~\mathrm{g} \\[1.5ex] &= 11.5~\mathrm{g} \end{align*}\]


Practice


Let’s say you mixed only 12.58 g of nitric acid with 10.00 g of magnesium hydroxide to produce water and magnesium nitrate.

  1. What is the limiting reactant and what is the theoretical yield (in g) of magnesium nitrate?
  2. If you obtained only 5.39 g of magnesium nitrate, what is the percent yield of the reaction?
  3. What mass (in g) of magnesium nitrate would need to be obtained to have a percent yield of 97.5 %?
Solution

Determine the limiting reactant by converting the amounts (in mol) of each reactant to amounts of a product using the stoichiometric factor. Here I will choose water as my product though you could choose magnesium nitrate for this.

\[\begin{align*} n(\mathrm{H_2O}) &= m(\mathrm{HNO_3}) ~ M(\mathrm{HNO_3})^{-1} ~ r(\mathrm{H_2O, HNO_3}) \\[1.5ex] &= 12.58~\mathrm{g} \left ( \dfrac{\mathrm{mol~HNO_3}}{148.32~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{2~mol~H_2O}}{\mathrm{2~mol~HNO_3}} \right ) \\[1.5ex] &= 0.0848\bar{1}6~\mathrm{mol} \\[3ex] n(\mathrm{H_2O}) &= m(\mathrm{Mg(OH)_2}) ~ M(\mathrm{Mg(OH)_2})^{-1} ~ r(\mathrm{H_2O, Mg(OH)_2}) \\[1.5ex] &= 12.58~\mathrm{g} \left ( \dfrac{\mathrm{mol~Mg(OH)_2}}{58.32~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{2~mol~H_2O}}{\mathrm{1~mol~Mg(OH)_2}} \right ) \\[1.5ex] &= 0.431\bar{4}1~\mathrm{mol} \end{align*}\]

We see that nitric acid is limiting since the amount going into the reaction can only produce about 0.085 mol of water whereas the available magnesium hydroxide could, in theory, produce a little over 0.4 mol of water.

Since nitric acid is the limiting reactant, use the amount of nitric acid to determine how much magnesium nitrate (in g) could be produced, in theory (i.e. the theoretical yield of magnesium nitrate).

\[\begin{align*} m(\mathrm{Mg(NO_3)_2}) &= m(\mathrm{HNO_3}) ~ M(\mathrm{HNO_3})^{-1} ~ r(\mathrm{Mg(NO_3)_2, HNO_3}) ~ M(\mathrm{Mg(NO_3)_2}) \\[1.5ex] &= 12.58~\mathrm{g} \left ( \dfrac{\mathrm{mol~HNO_3}}{148.32~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{1~mol~Mg(NO_3)_2}}{\mathrm{2~mol~HNO_3}} \right ) \left ( \dfrac{148.32~\mathrm{g}}{1~\mathrm{mol~Mg(NO_3)_2}} \right ) \\[1.5ex] &= 6.29\bar{0}0~\mathrm{g} \end{align*}\]

Next, find the percent yield of the reaction.

\[\begin{align*} \%~\mathrm{yield} &= \dfrac{\mathrm{actual~yield}}{\mathrm{theoretical~yield}} \times 100~\% \\[1.5ex] &= \dfrac{\mathrm{5.39~\mathrm{g}}}{\mathrm{6.29\bar{0}0~\mathrm{g}}} \times 100~\% \\[1.5ex] &= 85.\bar{6}9~\% \\[1.5ex] &= 85.7~\% \end{align*}\]

Finally, determine the mass of magnesium nitrate needed to have 97.5 % yield.

\[\begin{align*} \%~\mathrm{yield} &= \dfrac{\mathrm{actual~yield}}{\mathrm{theoretical~yield}} \times 100\% \longrightarrow \\[1.5ex] \mathrm{actual~yield} &= \left ( \dfrac{\%~\mathrm{yield}}{100~\%} \right ) \left ( \mathrm{theoretical~yield} \right ) \\[1.5ex] &= \left ( \dfrac{97.5~\%}{100~\%} \right ) \left ( 6.29\bar{0}0~\mathrm{g~Mg(NO_3)_2} \right ) \\[1.5ex] &= 6.1\bar{3}2~\mathrm{g} \\[1.5ex] &= 6.13~\mathrm{g~Mg(NO_3)_2} \end{align*}\]

Concentration

Concentration is a measure of the composition of a mixture and can be represented in many ways. A solution with a low concentration may be described as being dilute whereas a solution with a high concentration may be described as being concentrated.

Molar concentration, an intensive property, is a unit of concentration (c) defined by IUPAC as the “amount of constituent divided by the volume of the mixture”. It is often ascribed the units of moles of solute per liter of mixture or solution”.

\[c(X) = \dfrac{n(X)}{V(\mathrm{solution})}\]

IUPAC formally refers to molar concentration as “amount concentration” or “substance concentration”, both an abbreviation of “amount-of-substance concentration”. Molar concentration is also commonly referred to as molarity (M), an outdated word.

Molar concentration can be represented with square brackets. For example, to represent a 1.00 M aqueous NaCl solution, one could write

\[c(\mathrm{NaCl}) = [\mathrm{NaCl}] \longrightarrow \mathrm{mol~L^{-1}} = M\]


Practice


A solution is prepared by dissolving 83.5 g magnesium chloride (M = 95.21 g mol–1) in enough water to make a 500.0 mL solution. Find the molar concentrations (in mol L–1) of the following aqueous solutes:

  1. magnesium chloride
  2. magnesium(2+)
  3. chlorine(1–)
Solution

\[\begin{align*} \mathrm{MgCl(s)} &\longrightarrow \mathrm{Mg^{2+}(aq)} + 2~\mathrm{Cl^-(aq)} \end{align*}\]

[MgCl2]

Find the molar concentration (c) by dividing the amount of MgCl2 (in mol) by the total volume (in L) of the mixture.

\[\begin{align*} c(\mathrm{MgCl_2}) &= n(\mathrm{MgCl_2}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= m(\mathrm{MgCl_2}) ~ M(\mathrm{MgCl_2})^{-1} ~ V(\mathrm{solution})^{-1}\\[1.5ex] &= 83.5~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{95.21~\mathrm{g}} \right ) \left ( \dfrac{1}{500.0~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 1.7\bar{5}4~\mathrm{mol~L^{-1}} = 1.7\bar{5}4~M \end{align*}\]

[Mg2+]

This can be solved in two ways.

Find the molar concentration (c) by dividing the amount of Mg2+ by the total volume (in L) of the mixture. Find the amount of Mg2+ by using the mole ratio (r) with the moles of MgCl2.

\[\begin{align*} c(\mathrm{Mg^{2+}}) &= n(\mathrm{Mg^{2+}}) ~ V(\mathrm{solution})^{-1}\\[1.5ex] &= n(\mathrm{MgCl_2}) ~ r(\mathrm{Mg^{2+}, MgCl_2}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= m(\mathrm{MgCl_2}) ~ M(\mathrm{MgCl_2})^{-1} ~ r(\mathrm{Mg^{2+}, MgCl_2}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= 83.5~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{95.21~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{1~mol~Mg^{2+}}}{\mathrm{1~mol~MgCl_2}} \right ) \left ( \dfrac{1}{500.0~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 1.7\bar{5}4~\mathrm{mol~L^{-1}} = 1.7\bar{5}4~M \end{align*}\]

or by simply using the mole ratio with the molar concentration of MgCl2.

\[\begin{align*} c(\mathrm{Mg^{2+}}) &= c(\mathrm{MgCl_2}) ~ r(\mathrm{Mg^{2+}, MgCl_2}) \\[1.5ex] &=\dfrac{1.7\bar{5}4~\mathrm{moL~MgCl_2}}{\mathrm{L}} \left ( \dfrac{\mathrm{mol~Mg^{2+}}}{\mathrm{mol~MgCl_2}} \right ) \\[1.5ex] &= 1.7\bar{5}4~\mathrm{mol~L^{-1}} = 1.7\bar{5}4~M \end{align*}\]

[Cl]

This can be solved in two ways.

Find the molar concentration (c) by dividing the amount of Cl by the total volume (in L) of the mixture. Find the amount of Cl by using the mole ratio (r) with the moles of MgCl2. Note: You could also use the mole ratio with the moles of Mg2+ that we solved for above.

\[\begin{align*} c(\mathrm{Cl^-}) &= n(\mathrm{Cl^-})~ V(\mathrm{solution})^{-1} \\[1.5ex] &= n(\mathrm{MgCl_2}) ~ r(\mathrm{Cl^-, MgCl_2}) ~ V(\mathrm{solution})^{-1}\\[1.5ex] &= m(\mathrm{MgCl_2}) ~ M(\mathrm{MgCl_2})^{-1} ~ r(\mathrm{Cl^-, MgCl_2}) ~ V(\mathrm{solution})^{-1} \\[1.5ex] &= 83.5~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{95.21~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{2~mol~Cl^-}}{\mathrm{mol~MgCl_2}} \right ) \left ( \dfrac{1}{500.0~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[1.5ex] &= 3.5\bar{0}8~\mathrm{mol~L^{-1}} = 3.5\bar{0}8~M \end{align*}\]

or by simply using the mole ratio with the molar concentration of MgCl2 (or Mg2+).

\[\begin{align*} c(\mathrm{Cl^-}) &= c(\mathrm{MgCl_2}) ~ r(\mathrm{Cl^-, MgCl_2}) \\[1.5ex] &= 1.7\bar{5}4~M~\mathrm{MgCl_2} \left ( \dfrac{\mathrm{2~mol~Cl^-}}{\mathrm{mol~MgCl_2}} \right ) \\[1.5ex] &= 3.5\bar{0}8~\mathrm{mol~L^{-1}} = 3.5\bar{0}8~M \end{align*}\]


Dilution

There are two ways to make a solution less concentrated (or more dilute): (1) remove solute or (2) add solvent. Dilution is typically performed by increasing the amount of solvent present.

Consider a solution (“1”) with known molar concentration and volume. One can dilute the solution to achieve a desired concentration (2”). These quantities are related.

\[c_1V_1 = c_2V_2\]

For example, a 1.00 L aqueous stock solution with a molar concentration of 8.00 M must be diluted to a concentration of 0.875 M. How much water (in L) would need to be added?

First, find the volume of the solution after dilution (V2 in L).

\[\begin{align*} c(\mathrm{stock})~V(\mathrm{stock}) &= c(\mathrm{dilute})~V(\mathrm{dilute}) \longrightarrow \\[1.5ex] V(\mathrm{dilute}) &= c(\mathrm{stock})~V(\mathrm{stock})~c(\mathrm{dilute})^{-1} \\[1.5ex] &= \dfrac{8.00~\mathrm{mol~L^{-1}} (1.00~\mathrm{L})}{0.875~\mathrm{mol~L^{-1}}} \\[1.5ex] &= 9.1\bar{4}2~\mathrm{L} \end{align*}\]

Next, find the amount of water (in L) that was added for the dilution process.

\[\begin{align*} V(\mathrm{H_2O~added}) &= V(\mathrm{H_2O})_{\mathrm{dilute}} - V(\mathrm{H_2O})_{\mathrm{stock}} \\[1.5ex] &= 9.1\bar{4}2~\mathrm{L} - 1.00~\mathrm{L} \\[1.5ex] &= 8.1\bar{4}2~\mathrm{L} \end{align*}\]


Practice


A 50.00 mL aliquot of a 3.00 L 4.25 M aqueous stock solution is collected.

  1. What is the molar concentration of the 50.00 mL aliquot?
  2. If the aliquot is diluted by the addition of 15.00 mL of water, what is the final molar concentration of the dilute aliquot?
Solution

[solute]aliquot

Find the moles of solute in the aliquot.

\[\begin{align*} n(\text{solute})_{\mathrm{aliquot}} &= c(\mathrm{solute})_{\mathrm{stock}} ~ V(\mathrm{aliquot}) \\[1.5ex] &= \dfrac{4.25~\mathrm{mol}}{\mathrm{L}} ~ \left ( \dfrac{50.00~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \\[1.5ex] &= 0.21\bar{2}5~\mathrm{mol} \end{align*}\]

Next, find the molar concentration of the aliquot.

\[\begin{align*} c(\mathrm{solute})_{\mathrm{aliquot}} &= n(\mathrm{solute})_{\mathrm{aliquot}}~ V(\mathrm{solution})_{\mathrm{dilute}}^{-1} \\[1.5ex] &= 0.21\bar{2}5~\mathrm{mol} \biggr[ 50.00~\mathrm{mL} \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \biggr]^{-1} \\[1.5ex] &= 4.25~\mathrm{mol~L^{-1}} = 4.25~M \end{align*}\]

Notice how the molar concentration did not change. Molar concentration is an intensive property.

[solute]dilute

Find the molar concentration of the diluted aliquot by using the dilution equation.

\[\begin{align*} c(\mathrm{stock})~V(\mathrm{stock}) &= c(\mathrm{dilute})~V(\mathrm{dilute}) \longrightarrow \\[1.5ex] c(\mathrm{dilute}) &= c(\mathrm{stock})~V(\mathrm{stock}) ~ V(\mathrm{dilute})^{-1} \\[1.5ex] &= \dfrac{\dfrac{4.25~\mathrm{mol}}{\mathrm{L}} \left ( \dfrac{50.00~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right )} {\left ( \dfrac{(50.00 + 15.00)~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right )} \\[1.5ex] &= 3.2\bar{6}9~\mathrm{mol~L^{-1}} = 3.2\bar{6}9~M \end{align*}\]


Practice


A 28.63 g silver(I) nitrate sample is dissolved in enough water to make a 150.0 mL solution. Another solution is prepared by dissolving 11.50 g of potassium chloride in enough water to make a 150.0 mL solution. The solutions are mixed and a reaction is performed.

  1. If 21.1 g of precipitate was collected, what is the percent yield of reaction?
  2. What is the molar concentration of the aqueous product of reaction?
  3. What is the molar concentration of the precipitate?
Solution

\[\mathrm{AgNO_3(aq)} + \mathrm{KCl(aq)} \longrightarrow \mathrm{AgCl(s)} + \mathrm{KNO_3(aq)}\]

Find the percent yield

First find the limiting reactant.

\[\begin{align*} n(\mathrm{AgCl}) &= m(\mathrm{AgNO_3}) ~ M(\mathrm{AgNO_3})^{-1} ~ r(\mathrm{AgCl, AgNO_3}) \\[1.5ex] &= 28.63~\mathrm{g} \left ( \dfrac{\mathrm{mol~AgNO_3}}{169.87~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{1~mol~AgCl}}{\mathrm{1~mol~AgNO_3}} \right ) \\[1.5ex] &= 0.168\bar{5}4~\mathrm{mol} \\[3ex] n(\mathrm{AgCl}) &= m(\mathrm{KCl}) ~ M(\mathrm{KCl})^{-1} ~ r(\mathrm{AgCl, KCl}) \\[1.5ex] &= 11.50~\mathrm{g} \left ( \dfrac{\mathrm{mol~KCl}}{74.55~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{1~mol~AgCl}}{\mathrm{1~mol~KCl}} \right ) \\[1.5ex] &= 0.154\bar{2}5~\mathrm{mol} \\[3ex] \end{align*}\]

KCl is limiting. Next, find the theoretical yield of AgCl (in g).

\[\begin{align*} m(\mathrm{AgCl}) &= n(\mathrm{AgCl}) ~ M(\mathrm{AgCl}) \\[1.5ex] &= 0.154\bar{2}5~\mathrm{mol} \left ( \dfrac{143.32~\mathrm{g}}{\mathrm{mol}} \right ) \\[1.5ex] &= 22.1\bar{0}7~\mathrm{g} \end{align*}\]

Now find the percent yield.

\[\begin{align*} \%~\mathrm{yield} &= \dfrac{\mathrm{actual~yield}}{\mathrm{theoretical~yield}} \times 100\% \\[1.5ex] &= \dfrac{\mathrm{21.1~\mathrm{g}}}{\mathrm{22.1\bar{0}7~\mathrm{g}}} \times 100\% \\[1.5ex] &= 95.\bar{4}4~\% = 95.4~\% \end{align*}\]

Molar concentration of potassium nitrate

Find the theoretical yield of potassium nitrate.

\[\begin{align*} n(\mathrm{KNO_3}) &= n(\mathrm{AgCl}) ~ r(\mathrm{KNO_3, AgCl}) \\[1.5ex] &= 0.154\bar{2}5~\mathrm{mol} \left ( \dfrac{1~\mathrm{mol~KNO_3}}{1~\mathrm{mol~AgCl}} \right ) \\[1.5ex] &= 0.154\bar{2}5~\mathrm{mol} \end{align*}\]

Recall that the reaction only gave a 95.4 % yield. Find the actual yield of potassium nitrate

\[\begin{align*} \%~\mathrm{yield} &= \dfrac{\mathrm{actual~yield}}{\mathrm{theoretical~yield}} \times 100\% \longrightarrow \\[1.5ex] \mathrm{actual~yield} &= \left ( \dfrac{\%~\mathrm{yield}}{100~\%} \right ) ~ \mathrm{theoretical~yield} \\[1.5ex] &= \left ( \dfrac{95.\bar{4}4~\%}{100~\%} \right ) 0.154\bar{2}5~\mathrm{mol} \\[1.5ex] &= 0.147\bar{2}1~\mathrm{mol} \end{align*}\]

Next, find the volume (in L) of the final solution.

\[\begin{align*} V(\mathrm{solution}) &= V(\mathrm{AgNO_3}) + V(\mathrm{KCl}) \\[1.5ex] &= (150.0 + 150.0)~\mathrm{mL} \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \\[1.5ex] &= 0.300\bar{0}0~\mathrm{L} \end{align*}\]

Finally, determine the molar concentration of potassium nitrate.

\[\begin{align*} c(\mathrm{KNO_3}) &= n(\mathrm{KNO_3})~ V(\mathrm{solution})^{-1}\\[1.5ex] &= \dfrac{0.147\bar{2}1~\mathrm{mol}}{0.300\bar{0}0~\mathrm{L}} \\[1.5ex] &= 0.490\bar{7}0~\mathrm{mol~L^{-1}} = 0.4907~M \end{align*}\]

Molar concentration of silver chloride

Because silver chloride is effectively not soluble in water, the effective molar concentration is 0 M.