# To a solution of aluminum sulfate weighing 60.8 g and a mass fraction of 15% was added an excess of barium chloride solution.

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. Remember: The process of learning a person lasts a lifetime. The value of the same knowledge for different people may be different, it is determined by their individual characteristics and needs. Therefore, knowledge is always needed at any age and position.