Sodium chloride with a mass of 5.85 g was treated at room temperature with a solution of sulfuric

Sodium chloride with a mass of 5.85 g was treated at room temperature with a solution of sulfuric acid with a mass fraction of H2SO4 of 98% and a mass of 20 g, the released gas was passed through an excess of AgNO3 solution. Determine the mass of the resulting precipitate if the product yield in the first reaction is 80%.

Given:
m (NaCl) = 5.85 g
ω (H2SO4) = 98%
m (H2SO4) = 20 g
η1 = 80%

To find:
m (draft) -?

Decision:
1) NaCl + H2SO4 => NaHSO4 + HCl ↑;
HCl + AgNO3 => HNO3 + AgCl ↓;
2) n (NaCl) = m (NaCl) / M (NaCl) = 5.85 / 58.5 = 0.1 mol;
3) m (H2SO4) = ω (H2SO4) * m solution (H2SO4) / 100% = 98% * 20/100% = 19.6 g;
4) n (H2SO4) = m (H2SO4) / M (H2SO4) = 19.6 / 98 = 0.2 mol;
5) n theory. (HCl) = n (NaCl) = 0.1 mol;
6) n practical (HCl) = η (HCl) * n theory. (HCl) / 100% = 80% * 0.1 / 100% = 0.08 mol;
7) n (AgCl) = n practical. (HCl) = 0.08 mol;
8) m (AgCl) = n (AgCl) * M (AgCl) = 0.08 * 143.5 = 11.48 g.

Answer: The mass of AgCl is 11.48 g.