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 mole of P = 31.0 g 1 mole of O = 32.0 g Moles of P = 75.3 g / 31.0 g/mol = 2.43 mol Moles of O = 38.7 g / 32.0 g/mol = 1.21 mol Limiting reagent is O, since it makes only 1.21 moles of product, whereas P would have made 2.43 moles of product if it reacted completely. Therefore, 1.21 moles of product is the theoretical yield. Moles of P₄O₆ produced = 43.3 g / 219.9 g/mol = 0.197 mol (actual yield) Percent yield = (actual yield / theoretical yield) x 100 Percent yield = (0.197 / 1.21) x 100 = 16.3% Therefore, the best choice is B.

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

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