Reaction Yields

Question

Consider this reaction: 2 Al + 3 CuCl2 → 3 Cu + 2 AlCl3 The reaction begins with 29.54 g of Al and 71.48 g of copper II chloride. a) How many grams of Cu is produced? B) How many grams of AlCl3 is produced?

Answer 1

2Al + 3CuCl₂ ==> 3Cu + 2AlCl₃ ... balanced equation

Since we are given the amounts of BOTH reactants, the first thing we have to do is determine which reactant is limiting. An easy way to do this is to simply divide the MOLES of each reactant by the corresponding coefficient in the balanced equation.

For Al: 29.54 g Al x 1 mol Al / 26.98 g = 1.095 mols Al (÷2->0.55)

For CuCl₂: 71.48 g CuCl2 x 1 mol / 134.5 g = 0.5314 mols CuCl₂ (÷3->0.177)

Since 0.177 is less than 0.55, CuCl₂ is the limiting reactant, and we will now use the MOLES of CuCl₂ (0.5314), to determine the theoretical yield of Cu and AlCl₃ that can be formed.

Theoretical yield of products:

Use mols of limiting reactant and the mol ratio in the balanced equation, as follows:

0.5314 mols CuCl₂ x 3 mol Cu / 3 mol CuCl₂ x 63.55 g Cu / mol Cu = 33.77 g Cu

0.5314 mols CuCl₂ x 2 mol AlCl3 / 3 mol CuCl₂ x 133.3 g AlCl₃ / mol = 47.22 g AlCl₃