Calculate the mass of the precipitate that is formed by the interaction of 120 g of 78%

Calculate the mass of the precipitate that is formed by the interaction of 120 g of 78% sulfuric acid solution with barium nitrate.

1. Let’s write down the reaction equation:

Ba (NO3) 2 + H2SO4 = BaSO4 ↓ + 2HNO3;

2.Calculate the mass of sulfuric acid contained in the solution:

m (H2SO4) = w (H2SO4) * m (solution) = 0.78 * 120 = 93.6 g;

3. find the chemical amount of acid:

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

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

n (H2SO4) = 93.6: 98 = 0.9551 mol;

4.determine the amount of barium sulfate formed:

n (BaSO4) = n (H2SO4) = 0.9551 mol;

5.Calculate the mass of the sediment:

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

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

m (BaSO4) = 0.9551 * 233 = 222.5 g.

Answer: 222.5 g.