Question 3.14: Calculating Quantities of Reactants and Products: Amount (mo......
Calculating Quantities of Reactants and Products: Amount (mol) to Amount (mol)
Problem In a lifetime, the average American uses more than a half ton (>500 kg) of copper in coins, plumbing, and wiring. Copper is obtained from sulfide ores, such as chalcocite [copper(I) sulfide] by a multistep process. After initial grinding, the ore is “roasted” (heated strongly with oxygen gas) to form powdered copper(I) oxide and gaseous sulfur dioxide. How many moles of oxygen are required to roast 10.0 mol of copper(I) sulfide?
Plan We always write the balanced equation first. The formulas of the reactants are Cu_2S and O_2, and the formulas of the products are Cu_2O and SO_2, so we have
2Cu_2S(s) + 3O_2(g) ⟶ 2Cu_2O(s) + 2SO_2(g)We know the amount of Cu_2S (10.0 mol) and must find the amount (mol) of O_2 that is needed to roast it. The balanced equation shows that 3 mol of O_2 is needed to roast 2 mol of Cu_2S, so the conversion factor for finding amount (mol) of O_2 is 3 mol O_2/2 mol Cu_2S (see the road map (Fig 3.14)).
Solution Calculating the amount of O_2:
\text{Amount (mol) of }O_2 = 10.0 \cancel{\text{mol} Cu_2S} \times \frac{3 \text{mol} O_2}{2 \cancel{\text{mol} Cu_2S}} = 15.0 \text{mol} O_2Check The units are correct, and the answer is reasonable because this molar ratio of O_2 to Cu_2S (15/10) is identical to the ratio in the balanced equation (3/2).
Comment A common mistake is to invert the conversion factor; that calculation would be
\text{Amount (mol) of }O_2 = 10.0 \text{mol} Cu_2S \times \frac{2 \text{mol} Cu_2S}{3 \text{mol} O_2} = \frac{6.67 \text{mol}^2 Cu_2S}{1 \text{mol} O_2}The strange units should alert you that an error was made in setting up the conversion factor. Also note that this answer, 6.67, is less than 10.0, whereas the equation shows that there should be more moles of O_2 (3 mol) than moles of Cu_2S (2 mol). Be sure to think through the calculation when setting up the conversion factor and canceling units.
