Calculate the mass of the precipitate obtained by the interaction of barium nitrate

| September 17, 2021 | education

Calculate the mass of the precipitate obtained by the interaction of barium nitrate and 50 g of a sulfuric acid solution with a mass fraction of a solute of 9.6%

Chemical reaction equation
Ba (NO3) 2 + H2SO4 → BaSO4 ↓ + 2HNO3

Mass of sulfuric acid in solution:
m (H2SO4) = w (H2SO4) × mр-pa (H2SO4): 100 = 9.6 × 50: 100 = 4.8 g

Molar mass of sulfuric acid:
M (H2SO4) = 1 × 2 + 1 × 32 + 4 × 16 = 98 g / mol

The amount of sulfuric acid:
n (H2SO4) = m (H2SO4): M (H2SO4) = 4.8: 98 = 0.049 mol

From 1 mol of H2SO4, 1 mol of BaSO4 precipitate is formed. This means that 0.049 mol of H2SO4 forms 0.049 mol of BaSO4 precipitate.
Molar mass of barium sulfate BaSO4:
M (BaSO4) = 1 × 137 + 1 × 32 + 4 × 16 = 233 g / mol

Sludge weight:
m (BaSO4) = n (BaSO4) × M (BaSO4) = 0.049 × 233 = 11.42 g



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