Determine the volume of gas that will be released when dissolving copper weighing 6.4 g

Determine the volume of gas that will be released when dissolving copper weighing 6.4 g in a solution weighing 200 g with a mass fraction of nitrate acid of 20%.

Given:
m (Cu) = 6.4 g
m solution (HNO3) = 200 g
ω (HNO3) = 20%

To find:
V (gas) -?

1) 3Cu + 8HNO3 => 3Cu (NO3) 2 + 2NO ↑ + 4H2O;
2) M (Cu) = Ar (Cu) = 64 g / mol;
3) n (Cu) = m / M = 6.4 / 64 = 0.1 mol;
4) m (HNO3) = ω * m solution / 100% = 20% * 200/100% = 40 g;
5) 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;
6) n (HNO3) = m / M = 40/63 = 0.63 mol;
7) n (NO) = n (Cu) * 2/3 = 0.1 * 2/3 = 0.07 mol;
8) V (NO) = n * Vm = 0.07 * 22.4 = 1.6 liters.

Answer: The volume of NO is 1.6 liters.