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.)