How much chlorine is required to completely chlorinate a mixture of 3.2 g of copper and 2.8 g of iron?

Given:
m (Cu) = 3.2 g
m (Fe) = 2.8 g

To find:
V (Cl2) -?

1) Cu + Cl2 => CuCl2;
2Fe + 3Cl2 => 2FeCl3;
2) M (Cu) = Ar (Cu) = 64 g / mol;
3) n (Cu) = m (Cu) / M (Cu) = 3.2 / 64 = 0.05 mol;
4) n1 (Cl2) = n (Cu) = 0.05 mol;
5) M (Fe) = Ar (Fe) = 56 g / mol;
6) n (Fe) = m (Fe) / M (Fe) = 2.8 / 56 = 0.05 mol;
7) n2 (Cl2) = n (Fe) * 3/2 = 0.05 * 3/2 = 0.075 mol;
8) n total (Cl2) = n1 (Cl2) + n2 (Cl2) = 0.05 + 0.075 = 0.125 mol;
9) V (Cl2) = n (Cl2) * Vm = 0.125 * 22.4 = 2.8 liters.

Answer: The volume of Cl2 is 2.8 liters.