Hydrogen chloride HCl, obtained by reacting concentrated sulfuric acid H2SO4 with sodium

Hydrogen chloride HCl, obtained by reacting concentrated sulfuric acid H2SO4 with sodium chloride NaCl weighing 11.7 g, is passed through a solution of silver nitrate. This resulted in the formation of a precipitate weighing 20.09 g. Has all the table salt reacted?

Given:
m total (NaCl) = 11.7 g
m (sediment) = 20.09 g

Find:
m rest (NaCl) -?

Solution:
1) NaCl + H2SO4 => NaHSO4 + HCl ↑;
HCl + AgNO3 => HNO3 + AgCl ↓;
2) n (AgCl) = m (AgCl) / M (AgCl) = 20.09 / 143.5 = 0.14 mol;
3) n (NaCl) = n (HCl) = n (AgCl) = 0.14 mol;
4) m react. (NaCl) = n (NaCl) * M (NaCl) = 0.14 * 58.5 = 8.19 g;
5) m rest. (NaCl) = m total. (NaCl) – m reag. (NaCl) = 11.7 – 8.19 = 3.51 g;
6) Not all of the table salt has reacted.

Answer: Not all of the table salt has reacted, the mass of the remaining NaCl is 3.51 g.