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?

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.)

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