Cuso4(aq) + 2naoh(aq) cu(oh)2(s) + na2so4(aq)
ii. if 638.44 g cuso4 reacts with 240.0 naoh, which is the limiting reagent?
cuso4 = 638.44 g divided by 159.62 g/mol = about 4 moles
naoh = 240 g divided 40 g/mol = about 6 moles
limiting reagent is naoh correct?

Your answer is right.

Important elements to consider:

- to use the balanced equation (which you did)
- divide the masses of each compound by the correspondant molar masses (which you did)
- compare the theoretical proportions with the current proportions

Theoretical: 2 mol of Na OH : 1 mol of CuSO4
Then 4 mol of NaOH need 2 mol of CUSO4.

Given that you have more than 2 mol of of CUSO4 you have plenty of it and the NaOH will consume first, being this the limiting reagent.

You would do 1500/4 and the answer is 375

No because it not possibly