6.3: solution stoichiometry

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Learning Objectives

To solve quantitative problems involving the stoichiometry of reactions in solution.

Quantitative calculations involving reactions in solution are carried out with masses, however, volumes of solutions of known concentration are used to determine the number of moles of reactants. Whether dealing with volumes of solutions of reactants or masses of reactants, the coefficients in the balanced chemical equation give the number of moles of each reactant needed and the number of moles of each product that can be produced. An expanded version of the flowchart for stoichiometric calculations is shown in Figure $\PageIndex{1}$. The balanced chemical equation for the reaction and either the masses of solid reactants and products or the volumes of solutions of reactants and products can be used to determine the amounts of other species, as illustrated in the following examples.

Top: flowchart of steps to convert mass of one substance to moles to moles of another substance to volume using molarity. Bottom: flowchart of steps to convert volume of a substance to mass of another substance. — Figure $\PageIndex{1}$: An Expanded Flowchart for Stoichiometric Calculations. Either the masses or the volumes of solutions of reactants and products can be used to determine the amounts of other species in a balanced chemical equation.

The balanced chemical equation for a reaction and either the masses of solid reactants and products or the volumes of solutions of reactants and products can be used in stoichiometric calculations.

Example $\PageIndex{1}$ : Extraction of Gold

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:

\[ Zn(s) + 2[Au(CN)_2]^-(aq) \rightarrow [Zn(CN)_4]^{2-}(aq) + 2Au(s) \]

What mass of gold can be recovered 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:

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.
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; proceed to the stoichiometric calculation. Figure $\PageIndex{2}$ is adapted for this particular problem as follows:

Flowchart. Top: volume is converted to moles of A via the concentration. Moles of A is then converted to moles of B via the mole ratio. The molar mass of B is then used to convert moles of B to grams. Bottom: same flowchart as above, filled in with information given in the exercise.

As indicated in the strategy, 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{align} moles\: [Au(CN)_2 ]^-
& = V_L M_{mol/L} \\
& = 400 .0\: \cancel{L} \left( \dfrac{3 .30 \times 10^{4-}\: mol\: [Au(CN)_2 ]^-} {1\: \cancel{L}} \right) = 0 .132\: mol\: [Au(CN)_2 ]^- \end{align} $

B Because the coefficients of gold and the [Au(CN)₂]⁻ ion are the same in the balanced chemical equation, assuming 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]⁻ (i.e., 0.132 mol of Au). The problem asks for the mass of gold that can be obtained, so the number of moles of gold must be converted to the corresponding mass using the molar mass of gold:

$ \begin{align} mass\: of\: Au &= (moles\: Au)(molar\: mass\: Au) \\
&= 0 .132\: \cancel{mol\: Au} \left( \dfrac{196 .97\: g\: Au} {1\: \cancel{mol\: Au}} \right) = 26 .0\: g\: Au \end{align}$

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

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

Exercise $\PageIndex{1}$ : Lanthanum Oxalate

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

Summary

Either the masses or the volumes of solutions of reactants and products can be used to determine the amounts of other species in the balanced chemical equation. Quantitative calculations that involve the stoichiometry of reactions in solution use volumes of solutions of known concentration instead of masses of reactants or products. The coefficients in the balanced chemical equation tell how many moles of reactants are needed and how many moles of product can be produced.

Learning Objectives

Example \(\PageIndex{1}\) : Extraction of Gold

Exercise \(\PageIndex{1}\) : Lanthanum Oxalate

Summary