To a solution of aluminum sulfate weighing 60.8 g and a mass fraction of 15% was added an excess of barium chloride solution. Calculate the mass of the precipitate formed.
1) Let’s compose the reaction equation:
Al2 (SO4) 3 + 3BaCl2 = 3BaSO4 ↓ + 2AlCl3
2) Calculate the mass and amount of the substance of aluminum sulfate contained in the solution:
m (Al2 (SO4) 3) = (m (solution) ∙ ω) / 100% = (60.8 ∙ 15) / 100 = 9.12 g;
n (Al2 (SO4) 3) = m (Al2 (SO4) 3) / M (Al2 (SO4) 3) = 9.12 / 342 = 0.0267 mol.
3) Determine the mass of the sediment:
according to the reaction equation
n (BaSO4) = 3n (Al2 (SO4) 3) = 3 ∙ 0.0267 = 0.08 mol;
m (BaSO4) = n (BaSO4) ∙ M (BaSO4) = 0.08 ∙ 233 = 18.66 g.
Answer: 18.66 g.
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