Determine the mass of the precipitate that forms when 9.8 g of sulfuric acid reacts with barium nitrate.

1. Let us write down the equation of the reaction between sulfuric acid and barium nitrate:

Ba (NO3) 2 + H2SO4 = BaSO4 ↓ + 2HNO3;

2.Calculate the chemical amount of sulfuric acid:

n (H2SO4) = m (H2SO4): M (H2SO4);

M (H2SO4) = 2 + 32 + 4 * 16 = 98 g / mol;

n (H2SO4) = 9.8: 98 = 0.1 mol;

3. Determine the amount of the formed precipitate of barium sulfate:

n (BaSO4) = n (H2SO4) = 0.1 mol;

4.calculate the mass of sulfate:

m (BaSO4) = n (BaSO4) * M (BaSO4);

M (BaSO4) = 137 + 32 + 16 * 4 = 233 g / mol;

m (BaSO4) = 0.1 * 233 = 23.3 g.

Answer: 23.3 g.