Consider the balanced equation. cuso4 + zn > znso4 + cu if 200.0 g of copper(ii) sulfate react with an excess of zinc metal, what is the theoretical yield of copper? 1.253 g 50.72 g 79.63 g 194.3 g

Let's break this down. We know from our balanced equation that (in theory) we'll get the same number of moles of copper out of the reaction that we put into it. So we need to find the number of moles of CuSO4 we have in 200.0 grams. Using the molar mass of CuSO4:
200.0 grams CuSO4 * (1 mole CuSO4)/(159.61 grams CuSO4) =
1.253 moles CuSO4
We know that the moles of CuSO4 and Cu are one-to-one, so we should yield the same number of moles of copper. If we multiply by copper's molar mass, we get:
1.253 moles Cu * (63.55 grams Cu)/(1 mole Cu) = 79.63 grams Cu