Question

what is the percent yield of cus for the following reaction given that you start with 15.5 g of na₂s and 12.1 g cuso₄? the actual amount of cus produced was 3.05 g
reaction: na₂s + cuso₄ → na₂so₄ + cus
options: 7.25%, 42.1%, 18.93%, 16.1%, not enough information

Explanation:

Step1: Find limiting reactant

Molar mass of \( \ce{Na2S} \): \( 2\times23 + 32 = 78 \, \text{g/mol} \). Moles of \( \ce{Na2S} \): \( \frac{15.5\,\text{g}}{78\,\text{g/mol}} \approx 0.1987\,\text{mol} \).
Molar mass of \( \ce{CuSO4} \): \( 63.5 + 32 + 4\times16 = 159.5\,\text{g/mol} \). Moles of \( \ce{CuSO4} \): \( \frac{12.1\,\text{g}}{159.5\,\text{g/mol}} \approx 0.0759\,\text{mol} \).
Reaction: \( \ce{Na2S + CuSO4 -> Na2SO4 + CuS} \) (1:1 ratio). \( \ce{CuSO4} \) is limiting (fewer moles).

Step2: Calculate theoretical yield of \( \ce{CuS} \)

Molar mass of \( \ce{CuS} \): \( 63.5 + 32 = 95.5\,\text{g/mol} \).
Theoretical yield (from \( \ce{CuSO4} \)): \( 0.0759\,\text{mol} \times 95.5\,\text{g/mol} \approx 7.25\,\text{g} \).

Step3: Calculate percent yield

Percent yield \( = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 = \frac{3.05\,\text{g}}{7.25\,\text{g}} \times 100 \approx 42.1\% \).

Answer:

42.1% (corresponding to the option "42.1%")