Hydrogen chloride, which was obtained by reacting an excess of concentrated sulfuric acid with 58.5 g

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

Given:
m (NaCl) = 58.5 g
m (H2O) = 146 g

To find:
ω (HCl) -?

Decision:
1) H2SO4 + 2NaCl => Na2SO4 + 2HCl ↑;
2) n (NaCl) = m / M = 58.5 / 58.5 = 1 mol;
3) n (HCl) = n (NaCl) = 1 mol;
4) m (HCl) = n * M = 1 * 36.5 = 36.5 g;
5) m solution (HCl) = m (HCl) + m (H2O) = 36.5 + 146 = 182.5 g;
6) ω (HCl) = m * 100% / m solution = 36.5 * 100% / 182.5 = 20%.

Answer: Mass fraction of HCl is 20%.