c) Calculate the volume of sulfur dioxide produced when 460 g of copper pyrite, \( \mathrm{CuFeS}_{2} \) is used.

To find the volume of sulfur dioxide produced from 460 g of copper pyrite (\( \mathrm{CuFeS}_{2} \)), we start with the balanced chemical reaction for the combustion of copper pyrite: \[ \mathrm{4\ CuFeS_{2} + 9\ O_{2} \rightarrow 4\ Cu + 2\ Fe_{2}O_{3} + 8\ SO_{2}} \] 1. **Calculate the molar mass of \( \mathrm{CuFeS_2} \)**: \[ \text{Cu} = 63.55 \, \text{g/mol}, \quad \text{Fe} = 55.85 \, \text{g/mol}, \quad \text{S} = 32.07 \, \text{g/mol} \] \[ \text{Molar mass of } \mathrm{CuFeS_{2}} = 63.55 + 55.85 + (2 \times 32.07) = 183.47 \, \text{g/mol} \] 2. **Convert grams of copper pyrite to moles**: \[ \text{Moles of } \mathrm{CuFeS_{2}} = \frac{460 \, \text{g}}{183.47 \, \text{g/mol}} \approx 2.51 \, \text{mol} \] 3. **Use the stoichiometry of the reaction**: From the reaction, \( 4 \, \text{mol of } \mathrm{CuFeS_{2}} \) produces \( 8 \, \text{mol of } \mathrm{SO_{2}} \). Therefore: \[ \text{Moles of } \mathrm{SO_{2}} = 2.51 \, \text{mol CuFeS_{2}} \times \frac{8 \, \text{mol SO}_{2}}{4 \, \text{mol CuFeS_{2}}} = 5.02 \, \text{mol SO_{2}} \] 4. **Convert moles of \( \mathrm{SO_{2}} \) to volume at STP**: At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 L. \[ \text{Volume of } \mathrm{SO_{2}} = 5.02 \, \text{mol} \times 22.4 \, \text{L/mol} \approx 112.45 \, \text{L} \] Thus, the volume of sulfur dioxide produced is approximately 112.45 liters.