Determine the mass of salt and the amount of water substance obtained by the reaction

Determine the mass of salt and the amount of water substance obtained by the reaction of copper (II) hydroxide with 58.8 g of sulfuric acid.

Given:
m (H2SO4) = 58.8 g.
m (salt) -?
m (H2O) -?

Decision:
1) Cu (OH) 2 + H2SO4 = CuSO4 + 2H2O
2) n (H2SO4) = 58.8 / 98 = 0.6 mol
3) n (salts) = n (H2SO4) = 0.6 mol
4) m (CuSO4) = 0.6 * 160 = 96 g.
5) n (H2O) = 0.6 * 2 = 1.2 mol

Answer: 96 g, 1.2 mol