Question

If 45.0 g of Aluminum metal and 62.0 g of Chlorine gas are mixed, and 66.5g of aluminum chloride is formed, what is the percent yield of aluminum chloride for the reaction?

Answer 1

The balanced reaction is

2Al + 3Cl2 = 2AlCl3

2 moles or 53.96 grams of Al reacts with 3 moles or 212.7 grams of chlorine gas to produce 2 moles or 266.7 of AlCl3

45 grams of aluminium would require (45 x 212.7) / 53.96 = 177.4 grams of chlorine ,

so chlorine is the limiting reagent here .

Now 212.7 grams of chlorine makes 266.7 grams of aluminium chloride.

So 62.0 grams of chlorine will make ( 62.0 x 266.7 ) / 212.7 = 77.7 grams of aluminium chloride .

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

= ( 66.5 / 77.7 ) x 100

= 85.6 %