What is the percent yield for the reaction PCl₃(g) + Cl₂(g) → PCl₅(g)
If 119.3 g of PCl₅ (ℳ = 208.2 g/mol) are formed when 61.3 g of Cl₂ (ℳ = 70.91 g/mol) react with excess PCl₃?

Question

Quizplus · Accepted Answer

The first step is to determine the theoretical yield of PCl5 using stoichiometry:

1 mol PCl3 (137.33 g PCl3) reacts with 1 mol Cl2 (70.91 g Cl2) to produce 1 mol PCl5 (208.2 g PCl5).

This means that 61.3 g Cl2 will react with 2(61.3g Cl2 / 70.91 g/mol) = 1.724 moles of PCl3, which should produce:

1.724 moles PCl5 × 208.2 g/mol = 358.97 g PCl5 (theoretical yield).

Now, we can use the percent yield formula:

% yield = (actual yield/theoretical yield) x 100

% yield = (119.3 g PCl5 / 358.97 g PCl5) x 100 = 33.24%

Therefore, the percent yield is 66.3% (100% - 33.24%). The correct answer is choice C.

What Is the Percent Yield for the Reaction PCl3(g)

What Is the Percent Yield for the Reaction PCl₃(g)