Calculate the amount of substance and mass of copper (II) hydroxide, which can neutralize 12.6 g of nitric acid.

Given:
m (HNO3) = 12.6 g

To find:
n (Cu (OH) 2) -?
m (Cu (OH) 2) -?

1) Cu (OH) 2 + 2HNO3 => Cu (NO3) 2 + 2H2O;
2) M (HNO3) = Mr (HNO3) = Ar (H) * N (H) + Ar (N) * N (N) + Ar (O) * N (O) = 1 * 1 + 14 * 1 + 16 * 3 = 63 g / mol;
3) n (HNO3) = m / M = 12.6 / 63 = 0.2 mol;
4) n (Cu (OH) 2) = n (HNO3) / 2 = 0.2 / 2 = 0.1 mol;
5) M (Cu (OH) 2) = Mr (Cu (OH) 2) = Ar (Cu) * N (Cu) + Ar (O) * N (O) + Ar (H) * N (H) = 64 * 1 + 16 * 2 + 1 * 2 = 98 g / mol;
6) m (Cu (OH) 2) = n * M = 0.1 * 98 = 9.8 g.

Answer: The amount of substance Cu (OH) 2 is 0.1 mol; weight – 9.8 g.