Question 3.16: Calculating Quantities of Reactants and Products: Mass to Ma......
Calculating Quantities of Reactants and Products: Mass to Mass
Problem During the roasting of chalcocite (see Sample Problem 3.14), how many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide?
Plan In this problem, we know the mass of the product, Cu_2O (2.86 kg), and we need the mass (kg) of O_2 that reacts to form it. Therefore, we must convert from mass of product to amount of product to amount of reactant to mass of reactant. We convert the mass of Cu_2O from kg to g and then to amount (mol). Then, we use the molar ratio (3 mol O_2/2 mol Cu_2O) to find the amount (mol) of O_2 required. Finally, we convert the amount of O_2 to g and then kg (see the road map (Fig 3.16)).
Solution Converting from kilograms of Cu_2O to moles of Cu_2O: Combining the mass unit conversion with the mass-to-amount conversion gives
\text{Amount (mol) of }Cu_2O = 2.87 \cancel{\text{kg }Cu_2O} \times \frac{10^3 \cancel{g}}{1 \cancel{\text{kg}}} \times \frac{1 \text{mol} Cu_2O}{143.10 \cancel{g Cu_2O}} = 20.0 \text{mol} Cu_2OConverting from moles of Cu_2O to moles of O_2:
\text{Amount (mol) of }O_2 = 20.0 \cancel{\text{mol }Cu_2O} \times \frac{3 \text{mol} O_2}{2 \cancel{\text{mol} Cu_2O}} = 30.0 \text{mol} O_2Converting from moles of O_2 to kilograms of O_2: Combining the amount-to-mass conversion with the mass unit conversion gives
\text{Mass (kg) of }O_2 = 30.0 \cancel{\text{mol} O_2} \times \frac{32.00 g O_2}{1 \cancel{\text{mol} O_2}} \times \frac{1 \text{kg}}{10^3 \text{kg}} = 0.960 \text{kg} O_2Check The units are correct. Rounding to check the math, for example, in the final step, ∼30 mol × 30 g/mol × 1 kg/10³ g = 0.90 kg. The answer seems reasonable: even though the amount (mol) of O_2 is greater than the amount (mol) of Cu_2O, the mass of O_2 is less than the mass of Cu_2O because ℳ of O_2 is less than ℳ of Cu_2O.
Comment The three related sample problems (3.14–3.16)(3.14, 3.15) highlight the main point for solving stoichiometry problems: convert the information given into amount (mol). Then, use the appropriate molar ratio and any other conversion factors to complete the solution.
