Calculate the mass of the precipitate formed in the reaction of 71 g of sodium sulfate with a solution of barium chloride.

First, we write down the equation of reactions.
NaSO4 + BaCl2 = BaSO4 + 2NaCl.
First, we find the molar mass of sodium sulfate. We write down the solution.
M (Na2SO4) = 2 × 23 + 32 + 16 × 4 = 46 + 32 + 64 = 142 g / mol.
Next, we find the molar mass of barium sulfate, which precipitated.
M (BaSO4) = 233 g / mol.
Further, according to the reaction equation, we find the mass of the precipitate of barium sulfate.
71 g Na2SO4 – x g BaSO4.
142 g / mol Na2SO4 – 233 g / mol BaSO4.
Find the unknown value of x. We write down the solution.
X = 71 × 233 ÷ 142 = 116.5 g.
Answer: The mass of barium sulfate is 116.5 grams.