The gas obtained by the interaction of 9.52 g of copper with 50 ml of an 81% solution of nitric acid (p = 1.45 g / ml)

The gas obtained by the interaction of 9.52 g of copper with 50 ml of an 81% solution of nitric acid (p = 1.45 g / ml) was passed through 150 ml of a 20% solution of NaOH (p = 1.22 g / ml) Determine the mass fractions of the substances formed in the solution …

Given:
m (Cu) = 9.52 g
V solution (HNO3) = 50 ml
ω (HNO3) = 81%
ρ solution (HNO3) = 1.45 g / ml
V solution (NaOH) = 150 ml
ω (NaOH) = 20%
ρ solution (NaOH) = 1.22 g / ml

To find:
ω (substances) -?

1) Cu + 4HNO3 => Cu (NO3) 2 + 2NO2 + 2H2O;
2NO2 + 2NaOH => NaNO2 + NaNO3 + H2O;
2) n (Cu) = m / M = 9.52 / 64 = 0.149 mol;
3) m solution (HNO3) = ρ solution * V solution = 1.45 * 50 = 72.5 g;
4) m (HNO3) = ω * m solution / 100% = 81% * 72.5 / 100% = 58.725 g;
5) n (HNO3) = m / M = 58.725 / 63 = 0.932 mol;
6) n (NO2) = n (Cu) * 2 = 0.149 * 2 = 0.298 mol;
7) m (NO2) = n * M = 0.298 * 46 = 13.708 g;
8) m solution (NaOH) = ρ solution * V solution = 1.22 * 150 = 183 g;
9) m (NaOH) = ω * m solution / 100% = 20% * 183/100% = 36.6 g;
10) n (NaOH) = m / M = 36.6 / 40 = 0.915 mol;
11) n (NaNO2) = n (NO2) / 2 = 0.298 / 2 = 0.149 mol;
12) m (NaNO2) = n * M = 0.149 * 69 = 10.281 g;
13) n (NaNO3) = n (NO2) / 2 = 0.298 / 2 = 0.149 mol;
14) m (NaNO3) = n * M = 0.149 * 85 = 12.665 g;
15) m solution = m solution (NO2) + m solution (NaOH) = 13.708 + 183 = 196.708 g;
16) ω (NaNO2) = m * 100% / m solution = 10.281 * 100% / 196.708 = 5.23%;
17) ω (NaNO3) = m * 100% / m solution = 12.665 * 100% / 196.708 = 6.44%.

Answer: The mass fraction of NaNO2 is 5.23%; NaNO3 – 6.44%.