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 ( = 70.91 g/mol) react with excess PCl3?

Question

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 (

What is the percent yield for the reaction PCl<sub>3</sub>(g) + Cl<sub>2</sub>(g) → PCl<sub>5</sub>(g) If 119.3 g of PCl<sub>5</sub> (M = 208.2 g/mol) are formed when 61.3 g of Cl<sub>2</sub> ( = 70.91 g/mol) react with excess PCl3? A) 195% B) 85.0% C) 66.3% D) 51.4% E) 43.7%

= 70.91 g/mol) react with excess PCl3?

Quizplus · Accepted Answer

First, we need to calculate the theoretical yield of PCl₅: 1 mol Cl₂ reacts with 1 mol PCl3 to produce 1 mol PCl₅. Therefore, the number of moles of PCl₅ formed is: n = 61.3 g Cl₂/(70.91 g/mol Cl₂) = 0.864 mol Cl₂ Since the coefficients in the balanced equation are all 1, the number of moles of PCl₅ formed is also 0.864 mol. The mass of the actual yield of PCl₅ is given as 119.3 g. The percent yield is calculated as: percent yield = actual yield/theoretical yield x 100% percent yield = (119.3 g/0.864 mol x 208.2 g/mol) x 100% = 66.3% Therefore, the correct answer is C) 66.3%.

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

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