Question

Bornite (Cu3FeS3) is a copper ore used in the production of copper. When heated, the following reaction occurs. 2 Cu3FeS3(s) + 7 O2(g) 6 Cu(s) + 2 FeO(s) + 6 SO2(g) If 3.24 metric tons of bornite is reacted with excess O2 and the process has an 78.8% yield of copper, what mass of copper is produced?

Answer 1

$0.158 \; \text{tonnes}$

$2 \; \text{Cu}_3 \text{FeS}_3\; (s) + 7 \; \text{O}_2 \to 6 \; \text{Cu}\; (s) + 2\; \text{FeO}\; (s) + 6 \; \text{SO}_2 \; (g)$

Bornite $\text{Cu}_3 \text{FeS}_3$ has a molar mass of $342.65 \; \text{g} \cdot \text{mol}^{-1}$. $3.24 \; \text{tonnes}= 3.24 \times 10^{6} \; \text{g}$ of bornite would thus contain $3.24 \times 10^{6} / 342.65 \approx 9.46 \times 10^{3} \; \text{mol}$ formula units of this substance. Each two formula unit of bornite combusts to produce six formula units of copper metal $\text{Cu}$ as seen in the equation; $9.46 \times 10^{3} \; \text{mol}$ of bornite would therefore produce $2/6 \times 9.46 \times 10^{3} \; \text{mol} = 3.15 \times 10^{3} \; \text{mol}$ of copper, which corresponds to a theoretical yield of $m = n \cdot M = 3.15 \times 10^{3} \; \text{mol} \times 63.55 \; \text{g} \cdot \text{mol}^{-1} = 2.00 \times 10^{5} \; \text{g}$.

$\text{Actual Yield} = \text{Theoretical Yield} \times \text{Percentage Yield}$.

Thus the actual yield of copper: $2.00 \times 10^{5} \; \text{g} \times 78.8 \% = 1.58 \times 10^{5} \; \text{g} = 0.158 \; \text{tonnes}$