Question 3.4.2: Consider the reaction of gold with nitric acid and hydrochlo......
Consider the reaction of gold with nitric acid and hydrochloric acid.
Au + 3 HNO_{3} + 4 HCl → HAuCl_{4} + 3 NO_{2} + 3 H_{2}O
Determine the limiting reactant in a mixture containing 125 g of each reactant and calculate the maximum mass (in grams) of HAuCl_{4} that can be produced in the reaction.
You are asked to calculate the maximum mass of a product that can be formed in a reaction.
You are given the balanced equation and the mass of each reactant available.
In this case there are three reactants, so it is most efficient to calculate the maximum amount of product (HAuCl_{4}) that can be produced by each one. The reactant that produces the least amount of HAuCl_{4} is the limiting reactant
Step 1. Write the balanced chemical equation.
Au + 3 HNO_{3} + 4 HCl → HAuCl_{4} + 3 NO_{2} + 3 H_{2}O
Step 2. Use the molar mass of each reactant and the stoichiometric factors derived from the balanced equation to calculate the amount of HAuCl_{4} that can be produced by each reactant.
125 g Au\left(\frac{1\text{ mol Au}}{197.0\text{ g}} \right) \left(\frac{1\text{ mol HAuCl}_{4}}{1\text{ mol Au}} \right) \left(\frac{339.8\text{ g}}{1\text{ mol HAuCl}_{4}} \right) = 216 g HAuCl_{4}
125 g HNO_{3}\left(\frac{1\text{ mol HNO}_{3}}{63.01\text{ g}} \right) \left(\frac{1\text{ mol HAuCl}_{4}}{3\text{ mol HNO}_{3}} \right) \left(\frac{339.8\text{ g}}{1\text{ mol HAuCl}_{4}} \right) = 225 g HAuCl_{4}
125 g HCl\left(\frac{1\text{ mol HCl}}{36.46\text{ g}} \right) \left(\frac{1\text{ mol HAuCl}_{4}}{4\text{ mol HCl}} \right) \left(\frac{339.8\text{ g}}{1\text{ mol HAuCl}_{4}} \right) = 291 g HAuCl_{4}
Gold is the limiting reagent. The maximum amount of HAuCl_{4} that can be produced is 216 g.