Question

Consider the reaction between HCl and O2: 4HCl(g)+O2(g)→2H2O(l)+2Cl2(g) When 63.1 g of HCl are allowed to react with reactant17.2 g of O2, 55.0 g of Cl2 are collected. What is the limiting reactant? The theoretical yield of Cl2 for the reaction? The percent yield for the reaction?

Answer 1

4 moles or 145.84 grams of HCL react with 1 mole or 32 grams of O2 for complete reaction

So 63.1 g of HCL will require ( 63.1 x 32) / 145.84 = 13.84 grams of O2

So HCl is the limiting reagent as O2 is in excess.

If 145.84 grams of HC forms 2 moles or 141.812 grams of Cl2

then 63.1 g of HCl will produce ( 63.1 x 141.812) / 145.84

= 61.35 grams of Cl2

So the theoretical yield of Cl2 = 61.35 grams .

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

= ( 55.0 / 61.35) x 100

= 89.65 %