Question

2Al + 3CuS yields 3Cu + Al2S3 If you begin the reaction with 35.00 g of solid aluminum metal and 35.00 g of aqueous copper (II) sulfate, what is the maximum amount of copper metal that you can produce in grams?

Answer 1

Looks like the real equation is

$2Al + 3CuSO_4 => 3Cu + Al_2(SO_4)_3$

CuS -- is not copper (II) sulfate and is not soluble in water.

First find the amounts of reagents given to find out which reagent is in excess:

$$\nu = m/M$$

$\nu (Al) = 35/27 = 1.3 (mol)$

$\nu (CuSO_4) = 35/160 = 0.21875 (mol)$

As 2 mol of Al require 3 mol of CuSO4, then 1.3 mol of Al react with 1.95 mol of CuSO4, but we have less. So we calculate the mass of copper using CuSO4.

1 mol of CuSO4 contains 1 mol of Cu, then 0.21875 mol of CuSO4 contains 0.21875 mol of Cu.

$m = M * \nu$

$m(Cu) = 64 * 0.21875 = 14 (g)$

Answer: 14 g of copper metal we can produce if we begin the reaction with 35.00 g of solid aluminum metal and 35.00 g of aqueous copper (II) sulfate