What is the percent yield for the reaction PCl3(g) + Cl2(g) --> PCl5(g) if 119.3 g...

What is the percent yield for the reaction PCl₃(g) + Cl₂(g) --> PCl₅(g) if 119.3 g of PCl₅ are formed when 61.3 g of Cl₂ reacts with 125.5g PCl₃?

Expert Solution

Molar mass of PCL3 = 137.33 g/mol
mass of PCL3 = 125.5 g
number of moles of PCL3 = mass/molar mass = 125.5/137.33 = 0.914 mol

Molar mass of CL2 = 71 g/mol
mass of Cl2 = 61.3 g
number of moles of Cl2= mass/molar mass = 61.3/71=0.863 mol
So, Cl2 is limiting reagent.

theoretically,
number of moles of PCl5 produced = 0.863 mol
Molar mass of PCL5 = 208.24 g/mol
Mass of PCl5 produced = number of mole * molar mass = 0.863*208.24 =179.8 g

actual produced = 119.3 g
percentage yield = actual*100/theoretical
=119.3*100/179.8
=66.4%
Answer: 66.4 %

queen_honey_blossom answered 2 years ago

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