What is the theoretical yield of PI₃ from the reaction of 27.0 g of P and 68.0 g of I₂
2P(s) + 3I₂(s) $\rarr$ 2PI₃(s)

Question

What is the theoretical yield of PI₃ from the reaction of 27.0 g of P and 68.0 g of I₂ 2P(s) + 3I₂(s) $\rarr$2PI₃(s)

Quizplus · Accepted Answer

First, we need to balance the chemical equation: 2P(s) + 3I₂(s) → 2PI₃(s) Next, we need to determine the limiting reactant: 27.0 g P × (1 mol P/30.97 g P) = 0.871 mol P 68.0 g I₂ × (1 mol I₂/234.84 g I₂) = 0.290 mol I₂ The ratio of P to I₂ is 2:3, so I₂ is the limiting reactant. 0.290 mol I₂ × (1 mol PI₃/3 mol I₂) × (307.24 g PI₃/1 mol PI₃) = 73.5 g PI₃ Therefore, the theoretical yield of PI₃ is 73.5 g, which is closest to option C.

What Is the Theoretical Yield of PI3 from the Reaction →\rarr→

What Is the Theoretical Yield of PI₃ from the Reaction $\rarr$