The reaction between Na2S and CuSO4 will give us the balanced chemical reaction of,
Na2S + CUSO4 --> Na2SO4 + CuS
This means that for every 78g of Na2S, there needs to be 159.6 g of CuSO4. The ratio is equal to 0.4887 of Na2S: 1 of CuSO4. Thus, for every 12.1g of CuSO4, we need only 5.91 g of Na2S. Thus, there is an excess of 9.58 g of Na2S. The answer is letter C.
The is the B) doing background research about the question
We can rephrase the statement with a little more specificity in order to understand the answer here.
The mass of the products can never be more than the The mass that is expected.
