The amount of water substances is formed by the interaction of 5 mol of nitric acid with zinc hydroxide?

The amount of water substances is formed by the interaction of 5 mol of nitric acid with zinc hydroxide? Calculate the mass of one molecule of the resulting salt?

Given:
n (HNO3) = 5 mol

To find:
n (H2O) -?
m (1 mol. Salt) -?

1) Write the reaction equation:
2HNO3 + Zn (OH) 2 => Zn (NO3) 2 + 2H2O;
2) Determine the amount of substance H2O:
n (H2O) = n (HNO3) = 5 mol;
3) Calculate the molar mass of Zn (NO3) 2:
M (Zn (NO3) 2) = Mr (Zn (NO3) 2) = Ar (Zn) * N (Zn) + Ar (N) * N (N) + Ar (O) * N (O) = 65 * 1 + 14 * 2 + 16 * 6 = 189 g / mol;
4) m (1 mol Zn (NO3) 2) = M (Zn (NO3) 2) / NA = 189 / 6.02 * 10 ^ 23 = 3.14 * 10 ^ -22 g.

Answer: The amount of substance H2O is 5 mol; the mass of one Zn (NO3) 2 molecule is 3.14 * 10 ^ -22 g.