Determine the mass and quantity of the barium chloride substance to obtain a 2.5 molar precipitate of barium sulfate.

BaCl2 + H2SO4 = BaSO4 + 2HCl
n (BaSO4) = 2.5 mol
hence the amount of barium chloride substance = 2.5
mass of BaCl2 = (137 + 35.5 * 2) * 2.5 = 520 grams



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