Calculate the mass of precipitate formed by the interaction of an excess of barium chloride solution with a solution containing 10.26 g of aluminum sulfate

We compose the reaction equations
3 BaCl2 + Al2 (SO4) 3 = 3 BaSO4 + 2 AlCl3
Calculate the amount of aluminum sulfate substance n = 10.26 / 342 = 0.03 mol
Using the equation, calculate the amount of precipitate n (BaSO4) = 0.09 mol
m (BaSO4) = 0.09 * 233 = 20.97 g
Round up to the whole number 21g


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