Question

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?

Answer 1

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.