A copper ore consisting of 12.5% copper(II) sulfide, when heated and reacted with oxygen gas, produces copper(II) oxide and sulfur dioxide. If the reaction has a 90.0% yield, how many grams of copper(II) oxide are produced when 1.00 kg of the copper ore is processed?
A)93.7 g
B)104 g
C)116 g
D)150 g
E)749 g

Question

Quizplus · Accepted Answer

First, calculate the mass of copper(II) sulfide in 1.00 kg of the ore: $1.00 \, \text{kg} \times 0.125 = 125 \, \text{g}$. The reaction is: $\text{CuS} + \text{O}_2 \rightarrow \text{CuO} + \text{SO}_2$. Molar masses: $\text{CuS} = 95.61 \, \text{g/mol}$, $\text{CuO} = 79.55 \, \text{g/mol}$. Moles of $\text{CuS}$ in 125 g: $125 \, \text{g} / 95.61 \, \text{g/mol} = 1.307 \, \text{mol}$. Moles of $\text{CuO}$ produced (same as $\text{CuS}$ due to 1:1 ratio): $1.307 \, \text{mol}$. Mass of $\text{CuO}$ produced: $1.307 \, \text{mol} \times 79.55 \, \text{g/mol} = 103.96 \, \text{g}$. Considering 90% yield: $103.96 \, \text{g} \times 0.90 = 93.7 \, \text{g}$.

A Copper Ore Consisting of 12