Determine the mass of salt and the amount of water obtained by the interaction of copper hydroxide

Determine the mass of salt and the amount of water obtained by the interaction of copper hydroxide (2+) with 58.5 g of sulfuric acid.

Given:
m (H2SO4) = 58.5 g

To find:
m (salt) -?
n (H2O) -?

1) Cu (OH) 2 + H2SO4 => CuSO4 + 2H2O;
2) n (H2SO4) = m / M = 58.5 / 98 = 0.6 mol;
3) n (CuSO4) = n (H2SO4) = 0.6 mol;
4) m (CuSO4) = n * M = 0.6 * 160 = 96 g;
5) n (H2O) = n (H2SO4) * 2 = 0.6 * 2 = 1.2 mol.

Answer: The mass of CuSO4 is 96 g; the amount of H2O substance is 1.2 mol.