Question

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

Answer 1

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

Answer 2

Answer:

79.63 grams is the theoretical yield of copper.

Explanation:

$CuSO_4 + Zn\rightarrow ZnSO_4+Cu$

Moles of copper(II) sulfate = $\frac{200.0 g}{159.61 g/mol}=1.2530 mol$

According to recation, 1 mol of copper(II) sulfate  gives with 1 mole of copper metal .

Then 1.2530 moles of copper (II) sulfate will give :

$\frac{1}{1}\times 1.2530 mol=1.2530 mol$of copper metal.

Mass of 1.2530 moles of copper metal :

1.2530 mol × 63.546 g/mol = 79.63 g

79.63 grams is the theoretical yield of copper.