If 1.85 g al reacts with an excess copper(ii) sulphate and the percentage yield of cu is 56.6%, what mass of cu is produced

We have a balanced equation: 
2 Al+ 3 CuSO4 ⇒ Al2(SO4)3+ 3 Cu

1.85 g Al* (1 mol Al/ 26.98 g Al)* (3 mol Cu/ 2 mol Al)* ( 63.55 g Cu/ 1 mol Cu)= 6.54 g Cu.
(Note that the units cancel out so you get the answer)

6.54 g Cu* (56.6/100)= 3.70 g Cu.

The final answer is 3.70 g Cu~

Molar mass of :

Al = 27 g/moles

Cu = 63.5 g/moles

Balanced chemical equation:

3 CuSO4 + 2 Al = 1 Al2(SO4)3 + 3 Cu

2* 27 g Al > 3 * 63.5 g Cu
1.85 g Al > x g Cu

54 x =352.425

x = 352.425 / 54

x = 6.5263 g Cu

6.5263 g 100%
y g 56.6%

y = 56.63 . 6.5263 / 100

y = 369.38858 /100

y = 3.70 g of Cu