Tetraphosphorus hexaoxide ( = 219.9 g/mol) is formed by the reaction of phosphorus with oxygen gas. P₄(s) + 3O₂(g) $\to$ P₄O₆(s) If a mixture of 75.3 g of phosphorus and 38.7 g of oxygen produce 43.3 g of P₄O₆, what is the percent yield for the reaction? A) 57.5% B) 48.8% C) 38.0% D) 32.4% E) 16.3%

Question

Quizplus · Accepted Answer

1. Calculate the theoretical yield of P₄O₆: - Calculate moles of phosphorus and oxygen: Phosphorus: 75.3 g ÷ 30.97 g/mol = 2.43 mol Oxygen: 38.7 g ÷ 32 g/mol = 1.21 mol - Determine limiting reactant: Oxygen is limiting since it produces less product than the phosphorus. - Calculate the theoretical yield based on oxygen: Oxygen: 1.21 mol × (1 mol of P₄O₆/3 mol of oxygen) × 219.9 g/mol = 88.6 g 2. Calculate the percent yield: Percent yield = (actual yield/theoretical yield) × 100% Percent yield = (43.3 g/88.6 g) × 100% = 48.8% Therefore, the answer is B, 48.8%.

Tetraphosphorus Hexaoxide ( = 219 →\to→ P4O6(s) If a Mixture of 75

Tetraphosphorus Hexaoxide ( = 219 $\to$ P₄O₆(s)
If a Mixture of 75