Question

How many grams of the excess reactant remain assuming the reaction goes to completion and that you start with 15.5g of na2s and 12.1g cuso4?

Answer 1

First what is given to you?

15.5 g of Na_2S
12.1 g of CuSO_4

Then decipher your formula...

g of Na_2S/ g per mol of Na_2S/ 1 mol = mol of Na_2S
g of CuSO_4/ g per mol of CuSO_4/ 1 mol = mol of CuSO_4

This will tell you which will react entirely with the elements. Then you should be able to determine which has excess amounts.

Then with those answers--- ((mol of Na_2S)-(mol of CuSO_4)) * (g per mol of Na_2S) = left over.

(Hopefully that is clear. If not message me.)