What is the percent yield for a reaction in which 22.1 g Cu is isolated by reacting 45.5 g Zn with 70.1 g CuSO₄?  
Zn (s) + CuSO₄ (aq) → Cu (s) + ZnSO₄ (aq)

• A. 63.1 %
• B. 90.7 %
• C. 79.2 %
• D. 67.1 %
• E. 50.0 %

Answer: (C)

Verified step by step guidance

1. Identify the balanced chemical equation for the reaction: Zn (s) + CuSO4 (aq) → Cu (s) + ZnSO4 (aq). This equation shows that one mole of Zn reacts with one mole of CuSO4 to produce one mole of Cu.
2. Calculate the molar masses of the reactants and products. The molar mass of Zn is approximately 65.38 g/mol, CuSO4 is approximately 159.61 g/mol, and Cu is approximately 63.55 g/mol.
3. Determine the theoretical yield of Cu. First, calculate the moles of Zn and CuSO4 available: moles of Zn = 45.5 g / 65.38 g/mol, and moles of CuSO4 = 70.1 g / 159.61 g/mol. The limiting reactant will determine the theoretical yield.
4. Using the limiting reactant, calculate the theoretical yield of Cu in grams. The moles of the limiting reactant will be equal to the moles of Cu produced, as per the stoichiometry of the balanced equation.
5. Calculate the percent yield using the formula: Percent Yield = (Actual Yield / Theoretical Yield) * 100. The actual yield is given as 22.1 g of Cu. Substitute the actual yield and the calculated theoretical yield into the formula to find the percent yield.