Question Detail
Question
For the balanced equation shown below, if the reaction of 35.5 grams of ZnS produces 17.3 grams of SO₂, what is the percent yield? (Zn = 65.4 g, S = 32 g, O = 16 g). 2ZnS + 3O₂ → 2ZnO + 2SO₂
Correct Answer
Option C is the correct answer.
Detailed Explanation
Molar mass ZnS = 65.4+32 = 97.4 g/mol. Moles ZnS = 35.5/97.4 = 0.3645 mol. From stoichiometry, 2 mol ZnS → 2 mol SO₂, so theoretical moles SO₂ = 0.3645 mol. Molar mass SO₂ = 32+32=64 g/mol. Theoretical mass SO₂ = 0.3645 × 64 = 23.33 g. % yield = (17.3/23.33)×100% = 74.1% ≈ 74%. Option B is 74%.
Hint
% yield = (actual yield / theoretical yield) × 100%.