Calculate the mass of the precipitate formed when an excess of barium chloride solution interacts

Calculate the mass of the precipitate formed when an excess of barium chloride solution interacts with a solution containing 10.26 g of aluminum sulfate.

The reaction of interaction of aluminum sulfate with barium chloride is described by the following chemical reaction equation:
3BaCl2 + Al2 (SO4) 3 = 3BaSO4 + 2AlCl3;
When 3 mol of barium chloride and 1 aluminum sulfate interact, 3 mol of insoluble barium sulfate is synthesized.
Let’s calculate the chemical amount of a substance that is contained in 10.26 grams of aluminum sulfate.
M Al2 (SO4) 3 = 27 x2 + (32 + 16 x 4) x 3 = 342 grams / mol;
N Al2 (SO4) 3 = 10.26 / 342 = 0.03 mol;
Thus, 0.03 x 3 = 0.09 mol of barium sulfate will be synthesized.
Let’s calculate its weight.
M BaSO4 = 137 + 32 + 16 x 4 = 233 grams / mol;
m BaSO4 = 233 x 0.09 = 6.99 grams;