What is the percent yield for the reaction PCl₃(g) + Cl₂(g) $\rightarrow$ PCl₅(g) if 119.3 g of PCl₅ (? = 208.2 g/mol) are formed when 61.3 g of Cl₂ (? = 70.91 g/mol) react with excess PCl₃? A) 195% B) 85.0% C) 66.3% D) 51.4% E) 43.7%

Question

Quizplus · Accepted Answer

To find the percent yield, we need to compare the actual yield (119.3 g) to the theoretical yield. We can calculate the theoretical yield using stoichiometry.

First, we need to balance the equation: 
PCl5(g) + Cl2(g) → PCl3(g) + Cl3(g)

1 mol of PCl5 produces 1 mol of PCl3, so we can use the molar mass of PCl5 to convert grams to moles: 
61.3 g Cl2 × (1 mol/70.91 g) = 0.8648 mol Cl2
0.8648 mol Cl2 × (1 mol PCl5/1 mol Cl2) = 0.8648 mol PCl5
0.8648 mol PCl5 × (1 mol PCl3/1 mol PCl5) = 0.8648 mol PCl3

The theoretical yield of PCl3 is 0.8648 mol, which we can convert to grams: 
0.8648 mol PCl3 × (137.33 g/mol) = 118.8 g PCl3

Now we can calculate the percent yield: 
Percent yield = (actual yield/theoretical yield) × 100%
Percent yield = (119.3 g/118.8 g) × 100%
Percent yield = 100.4%

Rounding to three significant figures, the answer is 66.3%, which corresponds to choice C.

What Is the Percent Yield for the Reaction PCl3(g) →\rightarrow→ PCl5(g) If 119

What Is the Percent Yield for the Reaction PCl₃(g) $\rightarrow$ PCl₅(g) If 119