Question

Bornite (Cu3FS3) is a copper ore used in the production of copper.
When heated, the following (unbalanced) reaction occurs:
Cu3FS3(s) + O2(g) → Cu(s)+ FeO(s) + SO2(g)

If 2.50 kg of bornite is reacted with excess O2 and the process has an 82.5% yield of copper, how much copper is produced (in kg)?

Answer 1

Bornite (Cu3FS3) is a copper ore used in the production of copper.

When heated, the following (unbalanced) reaction occurs:

$$\mathrm{Cu3FS3(s)} + \mathrm{O2(g)} \rightarrow \mathrm{Cu(s)} + \mathrm{FeO(s)} + \mathrm{SO2(g)}$$

If $2.50\,\mathrm{kg}$ of bornite is reacted with excess O2 and the process has an $82.5\%$ yield of copper, how much copper is produced (in kg)?

## Answer

The full equation of bornite heating is the following:

$$2\mathrm{Cu_3FeS_3} + 7\mathrm{O_2} = 6\mathrm{Cu} + 2\mathrm{FeO} + 6\mathrm{SO_2}$$ $$\mathrm{n}(\mathrm{Cu_3FeS_3}) = (2.5 \times 10^3\,\mathrm{g}) / (64 \times 3 + 56 + 3 \times 32)\,\mathrm{g} \cdot \mathrm{mol^{-1}} = 7.27\,\mathrm{mol}$$ $$\mathrm{n}(\mathrm{Cu}) : \mathrm{n}(\mathrm{Cu_3FeS_3}) = 2 : 6 = 1 : 3$$ $$\mathrm{m}(\mathrm{Cu}) = 7.27\,\mathrm{mol} \cdot 3 \cdot 0.825 \cdot 64\,\mathrm{g} \cdot \mathrm{mol^{-1}} = 1151.6\,\mathrm{g} = 1.15\,\mathrm{kg}$$

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