How many grams?

Question

2 Al + 3 CuCl₂^.H₂O → 3 Cu + 2 AlCl₃ + 6 H₂O

Assuming 32.49 grams of Al are consumed in the presence of excess copper II chloride dihydrate, how many grams of AlCl₃ can be produced if the reaction will only produce 76.95 % yield?

Answer 1

Using stoichiometry and dimensional analysis...

32.49 g Al x 1 mol Al/26.98 g x 2 mol AlCl₃/2 mol Al x 133.3 g/mol = 160.5 g x 0.7695 = 123.5 g of AlCl₃