Calculate the mass of barium sulfate resulting from mixing a solution containing 20.8 g

Calculate the mass of barium sulfate resulting from mixing a solution containing 20.8 g of barium chloride with an excess of magnesium sulfate solution.

The reaction of barium chloride with magnesium sulfate is described by the following equation:
BaCl2 + MgSO4 = BaSO4 + MgCl2;
The entire volume of barium chloride will transform into an insoluble precipitate of barium sulfate, as barium chloride reacts with an excess of magnesium sulfate solution.
Find the molar masses of barium chloride and barium sulfate:
M BaCl2 = 137 x 35.5 x 2 = 208 grams / mol;
M BaSO4 = 137 + 32 + 16 x 4 = 233 grams / mol;
Find the amount of substance in 20.8 grams of barium chloride:
N BaCl2 = 20.8 / 208 = 0.1 mol;
Find the mass of 0.1 mol of barium sulfate:
M BaSO4 = 0.1 x 233 = 23.3 grams;