What is the limiting reagent when 27.0 g of P and 68.0 g of I₂ react according to the following chemical equation
2P(s) + 3I₂(s) $\rarr$ 2PI₃(s)

Question

What is the limiting reagent when 27.0 g of P and 68.0 g of I₂ react according to the following chemical equation 2P(s) + 3I₂(s) $\rarr$ 2PI₃(s)

Quizplus · Accepted Answer

To determine the limiting reagent, we need to calculate the amount of product that each reactant can produce, assuming they were the limiting reagent. Using the molar mass of P (30.97 g/mol) and I₂ (319.74 g/mol) and stoichiometry, we can calculate the theoretical yield of PI₃ from 27.0 g of P and 68.0 g of I₂: For P: n = m/M = 27.0 g / 30.97 g/mol = 0.871 mol Theoretical yield = 0.871 mol PI₃ For I₂: n = m/M = 68.0 g / 319.74 g/mol = 0.213 mol Theoretical yield = 0.426 mol PI₃ From these calculations, we can see that I₂ produces the lower amount of PI₃. Therefore, I₂ is the limiting reagent. The correct answer is B.

What Is the Limiting Reagent When 27 →\rarr→ 2PI3(s) A) P B) I2 C) PbI3 D) There Is

What Is the Limiting Reagent When 27 $\rarr$ 2PI₃(s)
A) P
B) I₂
C) PbI₃
D) There Is