Pure copper may be produced by the reaction of copper(I) sulfide with oxygen gas as follows:
Cu₂S(s) + O₂(g) → 2Cu(s) + SO₂(g)
If the reaction of 0.530 kg of copper(I) sulfide with excess oxygen produces 0.290 kg of copper metal,what is the percent yield?

Question

Quizplus · Accepted Answer

The percent yield is calculated using the formula: $\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$.First, convert the mass of copper(I) sulfide (Cu2S) to moles using its molar mass (Cu = 63.55 g/mol, S = 32.07 g/mol, thus Cu2S ≈ 159.16 g/mol): $0.530 \, \text{kg} = 530 \, \text{g}$, so moles of Cu2S = $530 \, \text{g} / 159.16 \, \text{g/mol} ≈ 3.33 \, \text{mol}$.From the balanced chemical equation, 1 mole of Cu2S produces 2 moles of Cu. Therefore, $3.33 \, \text{mol} \, \text{Cu2S} \times 2 = 6.66 \, \text{mol} \, \text{Cu}$.Convert moles of Cu to mass: $6.66 \, \text{mol} \times 63.55 \, \text{g/mol} ≈ 423.3 \, \text{g}$.Finally, calculate the percent yield: $\frac{290 \, \text{g}}{423.3 \, \text{g}} \times 100\% ≈ 68.5\%$.

Pure Copper May Be Produced by the Reaction of Copper(I)sulfide