Hydrogen chloride, obtained by the interaction of 58.5 g of sodium chloride with an excess of concentrated sulfuric acid

Hydrogen chloride, obtained by the interaction of 58.5 g of sodium chloride with an excess of concentrated sulfuric acid, was dissolved with 146 g of water. Calculate the mass fraction (%) of hydrogen chloride in the resulting solution.

Given:
m (NaCl) = 58.5 g
m (H2O) = 146 g
Find: ω (HCl) -?
Decision:
1) NaCl + H2SO4 => HCl + NaHSO4;
2) M (NaCl) = Mr (NaCl) = Ar (Na) * N (Na) + Ar (Cl) * N (Cl) = 23 * 1 + 35.5 * 1 = 58.5 g / mol;
M (HCl) = Mr (HCl) = Ar (H) * N (H) + Ar (Cl) * N (Cl) = 1 * 1 + 35.5 * 1 = 36.5 g / mol;
3) n (NaCl) = m (NaCl) / M (NaCl) = 58.5 / 58.5 = 1 mol;
4) n (HCl) = n (NaCl) = 1 mol;
5) m (HCl) = n (HCl) * M (HCl) = 1 * 36.5 = 36.5 g;
6) m solution (HCl) = m (HCl) + m (H2O) = 36.5 + 146 = 182.5 g;
7) ω (HCl) = m (HCl) * 100% / m solution (HCl) = 36.5 * 100% / 182.5 = 20%.
Answer: Mass fraction of HCl is 20%.