Three gases were mixed in a confined space and detonated. What is the fraction of the acid formed in this case

Three gases were mixed in a confined space and detonated. What is the fraction of the acid formed in this case, if the first gas was obtained as a result of the violent interaction of 34.2 g of calcium with water; the second gas is by thermal decomposition of 25.4 g of potassium nitrate; the third gas – as a result of electrolysis of 5.2 g of molten calcium chloride.

Given:
m (Ca) = 34.2 g
m (KNO3) = 25.4 g
m (CaCl2) = 5.2 g

To find:
ω (acid) -?

Decision:
1) Ca + 2H2O => Ca (OH) 2 + H2 ↑;
2KNO3 => 2KNO2 + O2 ↑;
CaCl2 => Ca + Cl2 ↑;
2) n (Ca) = m (Ca) / M (Ca) = 34.2 / 40 = 0.86 mol;
3) n (H2) = n (Ca) = 0.86 mol;
4) n (KNO3) = m (KNO3) / M (KNO3) = 25.4 / 101 = 0.25 mol;
5) n (O2) = n (KNO3) / 2 = 0.25 / 2 = 0.13 mol;
6) n (CaCl2) = m (CaCl2) / M (CaCl2) = 5.2 / 111 = 0.05 mol;
7) n (Cl2) = n (CaCl2) = 0.05 mol;
8) 2H2 + O2 => 2H2O;
H2 + Cl2 => 2HCl;
9) n (H2 on O2) = n (O2) * 2 = 0.13 * 2 = 0.26 mol;
10) n (H2 on Cl2) = n (Cl2) = 0.05 mol;
11) n (HCl) = n (Cl2) * 2 = 0.05 * 2 = 0.1 mol;
12) m (HCl) = n (HCl) * M (HCl) = 0.1 * 36.5 = 3.65 g;
13) n (H2O) = n (O2) * 2 = 0.13 * 2 = 0.26 mol;
14) m (H2O) = n (H2O) * M (H2O) = 0.26 * 18 = 4.68 g;
15) m solution (HCl) = m (HCl) + m (H2O) = 3.65 + 4.68 = 8.33 g;
16) ω (HCl) = m (HCl) * 100% / m solution (HCl) = 3.65 * 100% / 8.33 = 43.8%.

Answer: Mass fraction of HCl is 43.8%.