A mixture of two anhydrous salts of nitric acid weighing 31.28 g was fried in a closed vessel, the gases were evacuated

A mixture of two anhydrous salts of nitric acid weighing 31.28 g was fried in a closed vessel, the gases were evacuated and cooled. As a result, 15.48 g of nitric acid (w (HNO3) = 48.84%) and the remainder of nitrogen (I) oxide and oxygen gases were obtained. Determine the qualitative and quantitative composition of the initial mixture (the degree of conversion in each stage is taken as 100%).

Given:
m (Me (NO3) x; Me (NO3) y) = 31.28 g
m solution (HNO3) = 15.48 g
ω (HNO3) = 48.84%

Find:
Me (NO3) x -?
Me (NO3) y -?
m (Me (NO3) x) -?
m (Me (NO3) y) -?

1) N2O is formed only during the decomposition of NH4NO3:
NH4NO3 = (toC) => N2O + 2H2O;
2) 4NO2 + O2 + 2H2O => 4HNO3;
3) m (HNO3) = ω * m solution / 100% = 48.84% * 15.48 / 100% = 7.56 g;
4) n (HNO3) = m / M = 7.56 / 63 = 0.12 mol;
5) n (NO2) = n (HNO3) = 0.12 mol;
6) n2 (H2O) = n (HNO3) * 2/4 = 0.12 * 2/4 = 0.06 mol;
7) m3 (H2O) = m solution (HNO3) – m (HNO3) = 15.48 – 7.56 = 7.92 g;
8) n3 (H2O) = m3 / M = 7.92 / 18 = 0.44 mol;
9) n1 (H2O) = n2 (H2O) + n3 (H2O) = 0.06 + 0.44 = 0.5 mol;
10) n (NH4NO3) = n1 (H2O) / 2 = 0.5 / 2 = 0.25 mol;
11) m (NH4NO3) = n * M = 0.25 * 80 = 20 g;
12) m (Me (NO3) y) = m (Me (NO3) x; Me (NO3) y) – m (NH4NO3) = 31.28-20 = 11.28 g;
13) y = 1 (NO2 is not formed):
2MeNO3 = (toC) => 2MeNO2 + O2;
14) y = 2:
2Me (NO3) 2 = (toC) => 2MeO + 4NO2 + O2;
15) n (Me (NO3) 2) = n (NO2) * 2/4 = 0.12 * 2/4 = 0.06 mol;
16) M (Me (NO3) 2) = m / n = 11.28 / 0.06 = 188 g / mol;
17) Ar (Me) = Mr (Me (NO3) 2) – Ar (N) * N (N) – Ar (O) * N (O) = 188 – 14 * 2 – 16 * 3 * 2 = 64 – copper (Cu).

Answer: The mass of NH4NO3 is 20 g; Cu (NO3) 2 – 11.28 g.