Sulfuric acid weighing 20 g reacted with an excess of barium nitrate. Determine the mass of precipitated sediment.

Given:
m (H2SO4) = 20 g

To find:
m (draft) -?

Decision:
1) H2SO4 + Ba (NO3) 2 => BaSO4 ↓ + 2HNO3;
2) M (H2SO4) = Mr (H2SO4) = Ar (H) * N (H) + Ar (S) * N (S) + Ar (O) * N (O) = 1 * 2 + 32 * 1 + 16 * 4 = 98 g / mol;
3) n (H2SO4) = m (H2SO4) / M (H2SO4) = 20/98 = 0.2 mol;
4) n (BaSO4) = n (H2SO4) = 0.2 mol;
5) 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;
6) m (BaSO4) = n (BaSO4) * M (BaSO4) = 0.2 * 233 = 46.6 g.

Answer: The mass of BaSO4 is 46.6 g.