166.4 grams of a barium chloride solution was mixed with an excess of sodium sulfate and a precipitate

166.4 grams of a barium chloride solution was mixed with an excess of sodium sulfate and a precipitate weighing 46.6 grams was formed. Calculate the mass fraction in the barium chloride solution.

Given:
m solution (BaCl2) = 166.4 g
m (sediment) = 46.6 g

To find:
ω (BaCl2) -?

Decision:
1) BaCl2 + Na2SO4 => BaSO4 ↓ + 2NaCl;
2) M (BaCl2) = Mr (BaCl2) = Ar (Ba) * N (Ba) + Ar (Cl) * N (Cl) = 137 * 1 + 35.5 * 2 = 208 g / mol;
M (BaSO4) = Mr (BaSO4) = Ar (Ba) * N (Ba) + Ar (S) * N (S) + Ar (O) * N (O) = 137 * 1 + 32 * 1 + 16 * 4 = 233 g / mol;
3) n (BaSO4) = m (BaSO4) / M (BaSO4) = 46.6 / 233 = 0.2 mol;
4) n (BaCl2) = n (BaSO4) = 0.2 mol;
5) m (BaCl2) = n (BaCl2) * M (BaCl2) = 0.2 * 208 = 41.6 g;
6) ω (BaCl2) = m (BaCl2) * 100% / m solution (BaCl2) = 41.6 * 100% / 166.4 = 25%.

Answer: The mass fraction of BaCl2 is 25%.