Upon reaction of 1.274 g of copper sulfate with excess zinc metal, 0.392 g copper metal

Question

was obtained according to the equation:
CuSO4 (aq) + Zn(s) ⟶ Cu(s) + ZnSO4(aq)
What is the percent yield?

Answer 1

To calculate the percent yield of the reaction, we first need to determine the theoretical yield of copper metal (Cu) that can be obtained from the given amount of copper sulfate (CuSO₄).

1. Determine the moles of copper sulfate (CuSO₄):

   • The molar mass of CuSO₄ is calculated as follows:
     • Cu: 63.55 g/mol
     • S: 32.07 g/mol
     • O (4): 16.00 g/mol × 4 = 64.00 g/mol
     • Molar mass of CuSO₄ = 63.55 + 32.07 + 64.00 = 159.62 g/mol

   Using the mass of CuSO₄: \[ \text{Moles of CuSO₄} = \frac{\text{mass}}{\text{molar mass}} = \frac{1.274 \text{ g}}{159.62 \text{ g/mol}} \approx 0.008 \text{ moles} \]

2. From the balanced equation: \[ \text{CuSO₄ (aq)} + \text{Zn (s)} \rightarrow \text{Cu (s)} + \text{ZnSO₄ (aq)} \] The reaction shows a 1:1 molar ratio of CuSO₄ to Cu. Therefore, the moles of Cu produced will also be 0.008 moles.

3. Calculate the theoretical yield of copper (Cu):

   • The molar mass of copper (Cu) is:
     • Cu: 63.55 g/mol

   So, the theoretical yield (in grams) can be calculated as follows: \[ \text{Theoretical yield of Cu} = \text{moles of Cu} \times \text{molar mass of Cu} = 0.008 \text{ moles} \times 63.55 \text{ g/mol} \approx 0.5084 \text{ g} \]

4. Calculate the percent yield:

   • The actual yield of Cu obtained is 0.392 g. The percent yield is calculated using the formula: \[ \text{Percent yield} = \left(\frac{\text{actual yield}}{\text{theoretical yield}}\right) \times 100% \] Substituting the values: \[ \text{Percent yield} = \left(\frac{0.392 \text{ g}}{0.5084 \text{ g}}\right) \times 100% \approx 77.1% \]

The percent yield of the reaction is approximately 77.1%.