200 grams of a 4.9% sulfuric acid solution was poured into a solution containing an excess of barium nitrate.

200 grams of a 4.9% sulfuric acid solution was poured into a solution containing an excess of barium nitrate. Determine the mass of the precipitate obtained and the amount of nitric acid.

Given:
m solution (H2SO4) = 200 g
ω (H2SO4) = 4.9%

To find:
m (draft) -?
n (HNO3) -?

1) H2SO4 + Ba (NO3) 2 => BaSO4 ↓ + 2HNO3;
2) m (H2SO4) = ω * m solution / 100% = 4.9% * 200/100% = 9.8 g;
3) n (H2SO4) = m / M = 9.8 / 98 = 0.1 mol;
4) n (BaSO4) = n (H2SO4) = 0.1 mol;
5) m (BaSO4) = n * M = 0.1 * 233 = 23.3 g;
6) n (HNO3) = n (H2SO4) * 2 = 0.1 * 2 = 0.2 mol.

Answer: The mass of BaSO4 is 23.3 g; the amount of HNO3 substance is 0.2 mol.