This is a stoichiometry problem, where we have an initial amount of reactant and we need to find out how much of the product will we end up with, in order to do that we need to:
1. Set up the properly balanced equation, which the question already provided us
2 Cu + S -> Cu2S
2. See how many moles of reactant there are in the given amount of grams, but in this case we have 2 masses given in the question, which will directly lead us to think about limiting and excess reactant, even though the question did not ask us about it, we need to know it in order to find the right amount of product
Another thing important is molar ratio, which is 2:1 between Cu and S, this means that for each 2 moles of Cu we only need 1 mol of S in order to react
10.6g of Cu, and molar mass = 63.55g/mol
3.98g of S, and molar mass = 32.06g/mol
63.55g = 1 mol of Cu
10.6g = x moles of Cu
x = 0.17 moles of Cu
If we have 0.17 moles of Cu, we would need 0.08 moles of S in order to react
but we don't know if 3.98g of S is equal to 0.08 moles, let's check
32.06g = 1 mol of S
x grams = 0.08 moles of S
x = 2.56 grams of S, this means that we have more mass than we actually need to react, since we only need 2.56 grams and we have 3.98 grams, therefore Cu is the limiting reactant and S is in excess
Now we can find out how much product will be formed, using the number of moles of the limiting reactant, the molar ratio between Cu and Cu2S will be 2:1, same explanation as the one given between Cu and S, therefore we will end up with 0.08 moles of Cu2S as product
0.08 moles of Cu2S, molar mass = 159.16g/mol
159.16g = 1 mol of Cu2S
x grams = 0.08 moles of Cu2S
x = 12.73 grams of Cu2S