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 17.2 g of o2, 59.6 g of cl2 are collected.

The balanced chemical reaction is expressed as:

4hcl(g)+o2(g)→2h2o(l)+2cl2(g)

To determine the percent yield of the reaction, we need to calculate for the theoretical yield. This is the maximum amount of the product that can be produced from the reaction given the initial amounts of the reactants. First, we identify the limiting reactant as follows:

63.1 g of hcl ( 1 mol / 36.46 g ) ( 1 mol O2 / 4 mol HCl) ( 32 g / mol) = 13.85 g O2 
17.2 g of o2 ( 1 mol / 32 g ) ( 4 mol HCl / 1 mol O2) ( 36.46 g / mol) =  78.39 g HCl

Therefore, the limiting reactant would be HCl. We use the value for the HCl to calculate for the theoretical yield.

63.1 g of hcl ( 1 mol / 36.46 g ) ( 2 mol Cl2 / 4 mol HCl) ( 70.9 g / mol ) = 61.35 g Cl2

Percent yield = actual / theoretical x 100
                      = 59.6 / 61.35 x 100
                      = 97.1%

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