What is the answer to the equation

I'm still not good at stoichiometry, I need help on this equation

3CuCl2+2Al->2AlCl3+3Cu
Cu-Copper
Cl-Chlorine
Al-Aluminum

How many grams of Cu should be produced by using 15g of Al?

Suppose the above reaction produced 50g of Cu, What is the Percent Yield?

Can you show me step by step please?

3CuCl2 + 2Al --> 2AlCl3 + 3Cu
15g Al / 27g/mole = 0.556moles Al
0.556moles Al x (3Cu / 2Al) x 63.6g/mole = mass Cu = 53g

actual yield = 50g, theoretical yield = 53g
% yield = 50g / 53g x 100% = 94.34% yield