Example 7

Gold is extracted from its ores by treatment with an aqueous cyanide solution, which causes a reaction that forms the soluble [Au(CN)₂]⁻ ion. Gold is then recovered by reduction with metallic zinc according to the following equation:

Zn(s) + 2[Au(CN)₂]⁻(aq) → [Zn(CN)₄]²⁻(aq) + 2Au(s)

What mass of gold would you expect to recover from 400.0 L of a 3.30 × 10⁻⁴ M solution of [Au(CN)₂]⁻?

Given: chemical equation and molarity and volume of reactant

Asked for: mass of product

Strategy:

A Check the chemical equation to make sure it is balanced as written; balance if necessary. Then calculate the number of moles of [Au(CN)₂]⁻ present by multiplying the volume of the solution by its concentration.

B From the balanced chemical equation, use a mole ratio to calculate the number of moles of gold that can be obtained from the reaction. To calculate the mass of gold recovered, multiply the number of moles of gold by its molar mass.

Solution:

A The equation is balanced as written, so we can proceed to the stoichiometric calculation. We can adapt Figure 4.10 "An Expanded Flowchart for Stoichiometric Calculations" for this particular problem as follows:

As indicated in the strategy, we start by calculating the number of moles of [Au(CN)₂]⁻ present in the solution from the volume and concentration of the [Au(CN)₂]⁻ solution:

$$\begin{array}{ll}{\text{moles [Au(CN)}}_{\text{2}}{\text{]}}^{–}\hfill & =V_{\text{L}}{\text{M}}_{\text{mol/L}}\hfill \\ \hfill & =\text{ 400}\text{.0 }\cancel{\text{L}}\left(\frac{\text{3}\text{.30}\times {\text{10}}^{\text{4–}}{\text{ mol [Au(CN)}}_{\text{2}}{\text{]}}^{–}}{\text{1 }\cancel{\text{L}}}\right)=\text{ 0}{\text{.132 mol [Au(CN)}}_{\text{2}}{\text{]}}^{–}\hfill \end{array}$$

B Because the coefficients of gold and the [Au(CN)₂]⁻ ion are the same in the balanced chemical equation, if we assume that Zn(s) is present in excess, the number of moles of gold produced is the same as the number of moles of [Au(CN)₂]⁻ we started with (i.e., 0.132 mol of Au). The problem asks for the mass of gold that can be obtained, so we need to convert the number of moles of gold to the corresponding mass using the molar mass of gold:

$$\begin{array}{ll}\text{mass of Au }\hfill & =\text{ (moles Au)(molar mass Au)}\hfill \\ \hfill & =\text{ 0}\text{.132 }\cancel{\text{mol Au}}\left(\frac{\text{196}\text{.97 g Au}}{\text{1 }\cancel{\text{mol Au}}}\right)=\text{ 26}\text{.0 g Au}\hfill \end{array}$$

At a 2011 market price of over $1400 per troy ounce (31.10 g), this amount of gold is worth $1170.

$$\text{26}\text{.0 }\cancel{\text{g Au}}\times \frac{\text{1 }\cancel{\text{troy oz}}}{\text{31}\text{.10 }\cancel{\text{g}}}\times \frac{\text{\$1400}}{\text{1 }\cancel{\text{troy oz Au}}}=\text{ \$1170}$$

Exercise

What mass of solid lanthanum(III) oxalate nonahydrate [La₂(C₂O₄)₃·9H₂O] can be obtained from 650 mL of a 0.0170 M aqueous solution of LaCl₃ by adding a stoichiometric amount of sodium oxalate?

Answer: 3.89 g

Example 8

Because the consumption of alcoholic beverages adversely affects the performance of tasks that require skill and judgment, in most countries it is illegal to drive while under the influence of alcohol. In almost all US states, a blood alcohol level of 0.08% by volume is considered legally drunk. Higher levels cause acute intoxication (0.20%), unconsciousness (about 0.30%), and even death (about 0.50%). The Breathalyzer is a portable device that measures the ethanol concentration in a person’s breath, which is directly proportional to the blood alcohol level. The reaction used in the Breathalyzer is the oxidation of ethanol by the dichromate ion:

$${\text{3CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH(aq)}+\underset{\text{yellow-orange}}{{\text{2Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2}–}}\text{(aq)}+{\text{16H}}^{\text{ + }}\text{(aq)}\underset{{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)}}{\overset{{\text{Ag}}^{\text{ + }}}{\to }}\underset{\text{green}}{{\text{3CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H(aq)}+{\text{4Cr}}^{\text{3+}}\text{(aq)}+{\text{11H}}_{\text{2}}\text{O(l)}}$$

When a measured volume (52.5 mL) of a suspect’s breath is bubbled through a solution of excess potassium dichromate in dilute sulfuric acid, the ethanol is rapidly absorbed and oxidized to acetic acid by the dichromate ions. In the process, the chromium atoms in some of the Cr₂O₇²⁻ ions are reduced from Cr⁶⁺ to Cr³⁺. In the presence of Ag⁺ ions that act as a catalyst, the reaction is complete in less than a minute. Because the Cr₂O₇²⁻ ion (the reactant) is yellow-orange and the Cr³⁺ ion (the product) forms a green solution, the amount of ethanol in the person’s breath (the limiting reactant) can be determined quite accurately by comparing the color of the final solution with the colors of standard solutions prepared with known amounts of ethanol.

A typical Breathalyzer ampul contains 3.0 mL of a 0.25 mg/mL solution of K₂Cr₂O₇ in 50% H₂SO₄ as well as a fixed concentration of AgNO₃ (typically 0.25 mg/mL is used for this purpose). How many grams of ethanol must be present in 52.5 mL of a person’s breath to convert all the Cr⁶⁺ to Cr³⁺?

Given: volume and concentration of one reactant

Asked for: mass of other reactant needed for complete reaction

Strategy:

A Calculate the number of moles of Cr₂O₇²⁻ ion in 1 mL of the Breathalyzer solution by dividing the mass of K₂Cr₂O₇ by its molar mass.

B Find the total number of moles of Cr₂O₇²⁻ ion in the Breathalyzer ampul by multiplying the number of moles contained in 1 mL by the total volume of the Breathalyzer solution (3.0 mL).

C Use the mole ratios from the balanced chemical equation to calculate the number of moles of C₂H₅OH needed to react completely with the number of moles of Cr₂O₇²⁻ ions present. Then find the mass of C₂H₅OH needed by multiplying the number of moles of C₂H₅OH by its molar mass.

Solution:

A In any stoichiometry problem, the first step is always to calculate the number of moles of each reactant present. In this case, we are given the mass of K₂Cr₂O₇ in 1 mL of solution, which we can use to calculate the number of moles of K₂Cr₂O₇ contained in 1 mL:

$$\frac{{\text{moles K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}}{\text{1 mL}}=\frac{\text{(0}\text{.25 }\cancel{\text{mg}}{\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\text{)}}{\text{mL}}\left(\frac{\text{1 }\cancel{\text{g}}}{\text{1000 }\cancel{\text{mg}}}\right)\left(\frac{\text{1 mol}}{\text{294}\text{.18 }\cancel{\text{g}}{\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}}\right)=8.5\times {10}^{–7}\text{ moles}$$

B Because 1 mol of K₂Cr₂O₇ produces 1 mol of Cr₂O₇²⁻ when it dissolves, each milliliter of solution contains 8.5 × 10⁻⁷ mol of Cr₂O₇²⁻. The total number of moles of Cr₂O₇²⁻ in a 3.0 mL Breathalyzer ampul is thus

$${\text{moles Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2}–}=\left(\frac{\text{8}\text{.5}\times {\text{10}}^{–\text{7}}\text{ mol}}{\text{1 }\cancel{\text{mL}}}\right)(\text{3}\text{.0 }\cancel{\text{mL}})=\text{ 2}\text{.6}\times {\text{10}}^{–\text{6}}{\text{ mol Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{2–}$$

C The balanced chemical equation tells us that 3 mol of C₂H₅OH is needed to consume 2 mol of Cr₂O₇²⁻ ion, so the total number of moles of C₂H₅OH required for complete reaction is

$${\text{moles of C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH }=(\text{2}\text{.6}\times {\text{10}}^{–\text{6}}\cancel{{\text{mol Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2}–}})\left(\frac{{\text{3 mol C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}{\text{2 }\cancel{{\text{mol Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2}–}}}\right)=\text{ 3}\text{.9}\times {\text{10}}^{–\text{6}}{\text{ mol C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$$

As indicated in the strategy, this number can be converted to the mass of C₂H₅OH using its molar mass:

$${\text{mass C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH }=(\text{3}\text{.9}\times {\text{10}}^{–\text{6}}\cancel{{\text{mol C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}})\left(\frac{\text{46}\text{.07 g}}{\cancel{{\text{mol C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}}\right)=\text{ 1}\text{.8}\times {\text{10}}^{–\text{4}}{\text{ g C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$$

Thus 1.8 × 10⁻⁴ g or 0.18 mg of C₂H₅OH must be present. Experimentally, it is found that this value corresponds to a blood alcohol level of 0.7%, which is usually fatal.

Exercise

The compound para-nitrophenol (molar mass = 139 g/mol) reacts with sodium hydroxide in aqueous solution to generate a yellow anion via the reaction

Because the amount of para-nitrophenol is easily estimated from the intensity of the yellow color that results when excess NaOH is added, reactions that produce para-nitrophenol are commonly used to measure the activity of enzymes, the catalysts in biological systems. What volume of 0.105 M NaOH must be added to 50.0 mL of a solution containing 7.20 × 10⁻⁴ g of para-nitrophenol to ensure that formation of the yellow anion is complete?

Answer: 4.93 × 10⁻⁵ L or 49.3 μL

Example 9

When aqueous solutions of silver nitrate and potassium dichromate are mixed, an exchange reaction occurs, and silver dichromate is obtained as a red solid. The overall chemical equation for the reaction is as follows:

2AgNO₃(aq) + K₂Cr₂O₇(aq) → Ag₂Cr₂O₇(s) + 2KNO₃(aq)

What mass of Ag₂Cr₂O₇ is formed when 500 mL of 0.17 M K₂Cr₂O₇ are mixed with 250 mL of 0.57 M AgNO₃?

Given: balanced chemical equation and volume and concentration of each reactant

Asked for: mass of product

Strategy:

A Calculate the number of moles of each reactant by multiplying the volume of each solution by its molarity.

B Determine which reactant is limiting by dividing the number of moles of each reactant by its stoichiometric coefficient in the balanced chemical equation.

C Use mole ratios to calculate the number of moles of product that can be formed from the limiting reactant. Multiply the number of moles of the product by its molar mass to obtain the corresponding mass of product.

Solution:

A The balanced chemical equation tells us that 2 mol of AgNO₃(aq) reacts with 1 mol of K₂Cr₂O₇(aq) to form 1 mol of Ag₂Cr₂O₇(s) (Figure 4.11 "What Happens at the Molecular Level When Solutions of AgNO"). The first step is to calculate the number of moles of each reactant in the specified volumes:

$$\begin{array}{l}{\text{moles K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}=\text{ 500 }\cancel{\text{mL}}\left(\frac{\text{1 }\cancel{\text{L}}}{\text{1000 }\cancel{\text{mL}}}\right)\left(\frac{\text{0}{\text{.17 mol K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}}{\text{1 }\cancel{\text{L}}}\right)=\text{ 0}{\text{.085 mol K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\\ {\text{moles AgNO}}_{\text{3}}=\text{ 250 }\cancel{\text{mL}}\left(\frac{\text{1 }\cancel{\text{L}}}{\text{1000 }\cancel{\text{mL}}}\right)\left(\frac{\text{0}{\text{.57 mol AgNO}}_{\text{3}}}{\text{1 }\cancel{\text{L}}}\right)=\text{ 0}{\text{.14 mol AgNO}}_{\text{3}}\end{array}$$

B Now we can determine which reactant is limiting by dividing the number of moles of each reactant by its stoichiometric coefficient:

$$\begin{array}{l}{\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\text{: }\frac{\text{0}\text{.085 mol}}{\text{1 mol}}=\text{ 0}\text{.085}\\ {\text{AgNO}}_{\text{3}}\text{: }\frac{\text{0}\text{.14 mol}}{\text{2 mol}}=\text{ 0}\text{.070}\end{array}$$

Because 0.070 < 0.085, we know that AgNO₃ is the limiting reactant.

C Each mole of Ag₂Cr₂O₇ formed requires 2 mol of the limiting reactant (AgNO₃), so we can obtain only 0.14/2 = 0.070 mol of Ag₂Cr₂O₇. Finally, we convert the number of moles of Ag₂Cr₂O₇ to the corresponding mass:

$${\text{mass of Ag}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}=\text{ 0}\text{.070 }\cancel{\text{mol}}\left(\frac{\text{431}\text{.72 g}}{\text{1 }\cancel{\text{mol}}}\right)={\text{30 g Ag}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$$

Exercise

Aqueous solutions of sodium bicarbonate and sulfuric acid react to produce carbon dioxide according to the following equation:

2NaHCO₃(aq) + H₂SO₄(aq) → 2CO₂(g) + Na₂SO₄(aq) + 2H₂O(l)

If 13.0 mL of 3.0 M H₂SO₄ are added to 732 mL of 0.112 M NaHCO₃, what mass of CO₂ is produced?

Answer: 3.4 g