[tex]0.158 \; \text{tonnes}[/tex]
[tex]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)[/tex]
Bornite [tex]\text{Cu}_3 \text{FeS}_3[/tex] has a molar mass of [tex]342.65 \; \text{g} \cdot \text{mol}^{-1}[/tex]. [tex]3.24 \; \text{tonnes}= 3.24 \times 10^{6} \; \text{g}[/tex] of bornite would thus contain [tex]3.24 \times 10^{6} / 342.65 \approx 9.46 \times 10^{3} \; \text{mol}[/tex] formula units of this substance. Each two formula unit of bornite combusts to produce six formula units of copper metal [tex]\text{Cu}[/tex] as seen in the equation; [tex]9.46 \times 10^{3} \; \text{mol}[/tex] of bornite would therefore produce [tex]2/6 \times 9.46 \times 10^{3} \; \text{mol} = 3.15 \times 10^{3} \; \text{mol}[/tex] of copper, which corresponds to a theoretical yield of [tex]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}[/tex].
[tex]\text{Actual Yield} = \text{Theoretical Yield} \times \text{Percentage Yield}[/tex].
Thus the actual yield of copper: [tex]2.00 \times 10^{5} \; \text{g} \times 78.8 \% = 1.58 \times 10^{5} \; \text{g} = 0.158 \; \text{tonnes}[/tex]
