100 g of 10% sodium chloride solution reacts with silver nitrate solution. Calculate the mass of the precipitate formed.

Given:
m (NaCl) = 100 g
ω (NaCl) = 10%

To find:
m (draft) -?

Decision:
1) NaCl + AgNO3 => AgCl ↓ + NaNO3;
2) M (NaCl) = Mr (NaCl) = Ar (Na) * N (Na) + Ar (Cl) * N (Cl) = 23 * 1 + 35.5 * 1 = 58.5 g / mol;
M (AgCl) = Mr (AgCl) = Ar (Ag) * N (Ag) + Ar (Cl) * N (Cl) = 108 * 1 + 35.5 * 1 = 143.5 g / mol;
3) m (NaCl) = ω (NaCl) * m solution (NaCl) / 100% = 10% * 100/100% = 10 g;
4) n (NaCl) = m (NaCl) / M (NaCl) = 10 / 58.5 = 0.17 mol;
5) n (AgCl) = n (NaCl) = 0.17 mol;
6) m (AgCl) = n (AgCl) * M (AgCl) = 0.17 * 143.5 = 24.4 g.

Answer: The mass of AgCl is 24.4 g.