200 g of a 30% barium sulfate solution was treated with a sulfuric acid solution. Determine the mass of the resulting sediment.

| January 8, 2021 | education

An error was made in the condition. For barium sulfate when treated with sulfuric acid will give nothing. This is the precipitate of this acid.
Therefore, it would be wiser to put the condition like this (and in school problems they take this very substance): 200 g of a 30% solution of barium CHLORIDE was treated with a solution of sulfuric acid.
BaCl2 + H2SO4 = BaSO4 (precipitate, barium sulfate) + 2HCl.
Weight of pure barium chloride:
m (BaCl2) = 200 * 0.3 = 60 g.
Barium Chloride Amount:
n (BaCl2) = 60 / 208.246 = 0.2881 mol.
From the stoichiometric coefficients in the reaction equation, we see that the amount of chloride and barium sulfate is equal. Therefore:
n (BaSO4) = 0.2881 mol.
The mass of this sediment:
m (BaSO4) = n * m = 0.2881 * 233.38 = 67.238 g.



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