Calculate the mass of the salt formed by the interaction of 210 g of a 15% nitric acid solution

Calculate the mass of the salt formed by the interaction of 210 g of a 15% nitric acid solution with a sufficient amount of copper oxide (2)

Given:
m solution (HNO3) = 210 g
ω (HNO3) = 15%

To find:
m (salt) -?

1) 2HNO3 + CuO => Cu (NO3) 2 + H2O;
2) m (HNO3) = ω (HNO3) * m solution (HNO3) / 100% = 15% * 210/100% = 31.5 g;
3) 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;
4) n (HNO3) = m (HNO3) / M (HNO3) = 31.5 / 63 = 0.5 mol;
5) n (Cu (NO3) 2) = n (HNO3) / 2 = 0.5 / 2 = 0.25 mol;
6) M (Cu (NO3) 2) = Mr (Cu (NO3) 2) = Ar (Cu) * N (Cu) + Ar (N) * N (N) + Ar (O) * N (O) = 64 * 1 + 14 * 2 + 16 * 6 = 188 g / mol;
7) m (Cu (NO3) 2) = n (Cu (NO3) 2) * M (Cu (NO3) 2) = 0.25 * 188 = 47 g.

Answer: The mass of Cu (NO3) 2 is 47 g.