Determine the mass of the precipitate that forms when 9.8 g of sulfuric acid reacts with barium nitrate.
July 27, 2021 | education|
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.
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