284 g of sodium sulfate solution with a mass fraction of solute of 10% was mixed with an excess of barium nitrate solution.

284 g of sodium sulfate solution with a mass fraction of solute of 10% was mixed with an excess of barium nitrate solution. Calculate the mass of the precipitated sediment.

Given:
m solution (Na2SO4) = 284 g
ω (Na2SO4) = 10%

To find:
m (draft) -?

Decision:
1) Na2SO4 + Ba (NO3) 2 => 2NaNO3 + BaSO4 ↓;
2) M (Na2SO4) = Mr (Na2SO4) = Ar (Na) * N (Na) + Ar (S) * N (S) + Ar (O) * N (O) = 23 * 2 + 32 * 1 + 16 * 4 = 142 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) m (Na2SO4) = ω (Na2SO4) * m solution (Na2SO4) / 100% = 10% * 284/100% = 28.4 g;
4) n (Na2SO4) = m (Na2SO4) / M (Na2SO4) = 28.4 / 142 = 0.2 mol;
5) n (BaSO4) = n (Na2SO4) = 0.2 mol;
6) m (BaSO4) = n (BaSO4) * M (BaSO4) = 0.2 * 233 = 46.6 g.

Answer: The mass of BaSO4 is 46.6 g.