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

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Chemistry

What is the percent yield for the reaction PCl3(g)+ Cl2(g)--> PCl5(g)  if 119.3 g of PCl5are formed when 61.3 g of Cl2reacts with 125.5g PCl3?

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4.3 Ratings (777 Votes)

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 %
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