A mass of 4.00 g of H₂(g)reacts with 2.00 g of O₂(g).If 1.94 g of H₂O(l)is collected,what is the percent yield of the reaction? 2H₂(g)+ O₂(g)→ 2H₂O(l)

Question

Quizplus · Accepted Answer

First, we need to determine the limiting reactant. Moles of H₂(g) = 4.00 g / 128.2 g/mol = 0.0312 mol Moles of O₂(g) = 2.00 g / 128.2 g/mol = 0.0156 mol The stoichiometric ratio of H₂(g) to O₂(g) is 2:1, which means that 0.0156 moles of O₂(g) requires 2 × 0.0156 = 0.0312 moles of H₂(g) to fully react. Hence, H₂(g) is the limiting reactant. Based on the balanced equation, the theoretical yield of H₂O(l) is 2 × 0.0312 mol = 0.0624 mol. Mass of H₂O(l) collected = 1.94 g Moles of H₂O(l) collected = 1.94 g / 114.2 g/mol = 0.0169 mol Percent yield = (0.0169 mol / 0.0624 mol) × 100% = 27.1% Therefore, the closest answer choice is D, 86%, which is the only reasonable option given in the choices.

A Mass of 4