Mixed 52.6 ml of a 25% solution of aluminum sulfate (pl. 1.3 g / ml) and a solution of barium

| September 3, 2021 | education

Mixed 52.6 ml of a 25% solution of aluminum sulfate (pl. 1.3 g / ml) and a solution of barium chloride. Calculate the mass of the precipitate formed.

1. Al2 (SO4) 3 + 3BaCl2 = 3BaSO4 + 2AlCl3;

2.m (solution Al2 (SO4) 3) = ρ (solution Al2 (SO4) 3) * V (solution Al2 (SO4) 3);

m (solution Al2 (SO4) 3) = 1.3 * 52.6 = 68.38 g;

3.m (Al2 (SO4) 3) = w (Al2 (SO4) 3) * m (solution Al2 (SO4) 3);

m (Al2 (SO4) 3) = 0.25 * 68.28 = 17.095 g;

4.find the chemical amount of aluminum sulfate:

n (Al2 (SO4) 3) = m (Al2 (SO4) 3): M (Al2 (SO4) 3);

M (Al2 (SO4) 3) = 2 * 27 + 3 * 32 + 3 * 4 * 16 = 342 g / mol;

n (Al2 (SO4) 3) = 17.095: 342 = 0.05 mol

5.According to the reaction equation, the amount of precipitate obtained is three times greater than the consumed aluminum sulfate:

n (BaSO4) = 3 * n (Al2 (SO4) 3) = 3 * 0.05 = 0.15 mol;

6.Calculate the mass of the sediment:

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

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

m (BaSO4) = 0.15 * 233 = 34.95 g.

Answer: 34.95 g.



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