Theoretical yield question please help

Question

Calculate the theoretical yield for this experiment knowing that you are reacting 2.00 grams of CuCl2*2H2O (MM = 170.5 g/mol) with a certain mass of aluminum. The reaction is represented by: 

3 CuCl2●2H2O(aq) + 2 Al(s)  ----->  3 Cu(s) + 2AlCl3(aq) + 6 H2O(l)

Answer 1

This is kind of a poorly worded question, but here’s what I’m going to assume:

1) The question is saying there is an “excess” amount of aluminum and aluminum is not our limited reagent.

2) We will be calculating the theoretical yield of ALL products (because it doesn’t say which one, so we will just do all).

3) The question would like the theoretical yield represented in grams instead of moles (very common, but not specifically defined, here).

With that in mind, and the fact that we are given the chemical equation already balanced, an outline of what we will be doing is:

Step 1) convert reactant(s) to moles

Step 2) use ratio of balanced chemical equation to find moles of products from moles of reactant

Step 3) convert moles of products to grams of products using atomic mass of product

3 CuCl2●2H2O(aq) + 2 Al(s)  ----->  3 Cu(s) + 2AlCl3(aq) + 6 H2O(l)

Step 1

(2 g CuCl2●2H2O)(1 mole / 170.5 g) = 0.012 mole CuCl2●2H2O

Step 2

(0.012 mole CuCl2●2H2O)(3 Cu / 3 CuCl2●2H2O) = 0.012 mole Cu

(0.012 mole CuCl2●2H2O)(2 AlCl3 / 3 CuCl2●2H2O) = 0.008 mol AlCl3

(0.012 mole CuCl2●2H2O)(6 H2O / 3 CuCl2●2H2O) = 0.024 mol H2O

Step 3

(0.012 mole Cu)(63.5 g / 1 mol) = 0.762 g Cu

(0.008 mol AlCl3)(133.34 g / 1 mol) = 1.067 g AlCl3

(0.024 mol H2O)(18 g / 1 mol) = 0.432 g H2O