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Multiple Choice
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 %
Verified step by step guidance
1
Identify the balanced chemical equation for the reaction: Zn (s) + CuSO4 (aq) → Cu (s) + ZnSO4 (aq). This indicates a 1:1 molar ratio between Zn and Cu.
Calculate the moles of Zn used: Use the molar mass of Zn (approximately 65.38 g/mol) to convert 45.5 g of Zn to moles.
Calculate the moles of CuSO4 available: Use the molar mass of CuSO4 (approximately 159.61 g/mol) to convert 70.1 g of CuSO4 to moles.
Determine the limiting reactant: Compare the moles of Zn and CuSO4 to find which reactant is limiting, based on the 1:1 molar ratio.
Calculate the theoretical yield of Cu: Use the moles of the limiting reactant to find the theoretical moles of Cu produced, then convert this to grams using the molar mass of Cu (approximately 63.55 g/mol). Finally, calculate the percent yield using the formula: (actual yield / theoretical yield) * 100%.