Question

In the production of iron metal, a mixture of 228g P4 (MW=154 g/mol) and 205g Cl2 (MW=71 g/mol) are heated together the following reaction takes place to form PCl5 (MW=208 g/mol): P4(g)+10 Cl2(g)---> 4PCl5(g).

a) what is the limiting reactant?

b) How many grams of PCl5 can be produced?

c) If 157 g of Pcl5 is actually obtained what is the percent yield of the reaction?

Answer 1

Ans

a) 228 g of P4 = 228 / 154 = 1.48 moles

205 g of Cl2 = 205 / 71 = 2.89 moles

1 mole of P4 requires 10 moles of Cl2

So 1.48 moles of P4 will require 1.48 x 10 = 14.8 moles of Cl2

But we have less Cl2 present , so Cl2 is the limiting reagent.

b) 10 moles of Cl2 makes 4 moles of PCl5

So 2.89 moles of Cl2 will make ( 2.89 x 4) / 10 = 1.156 moles of PCl5

The amount of PCl5 = 1.156 x 208 = 240.45 grams

c) Percent yield = (actual yield / theoretical yield ) x 100

= (157 / 240.45) x 100

= 65.3 %