Calculate the mass of the salt formed by the reaction of 80 g of copper (II) oxide with hydrochloric acid.

Given:
m (CuO) = 80 g

To find:
m (salt) -?

1) CuO + 2HCl => CuCl2 + H2O;
2) M (CuO) = Mr (CuO) = Ar (Cu) * N (Cu) + Ar (O) * N (O) = 64 * 1 + 16 * 1 = 80 g / mol;
3) n (CuO) = m / M = 80/80 = 1 mol;
4) n (CuCl2) = n (CuO) = 1 mol;
5) M (CuCl2) = Mr (CuCl2) = Ar (Cu) * N (Cu) + Ar (Cl) * N (Cl) = 64 * 1 + 35.5 * 2 = 135 g / mol;
6) m (CuCl2) = n * M = 1 * 135 = 135 g.

Answer: The mass of CuCl2 is 135 g.