An excess of sodium sulfate solution was added to a solution containing 2.08 g

An excess of sodium sulfate solution was added to a solution containing 2.08 g of barium chloride. Determine the mass of the precipitated sediment.

We draw up the reaction equation and place the coefficients: BaCl2 + Na2SO4 = BaSO4 + 2NaCl.
The precipitate is barium sulfate.
1) Find the molar mass of barium chloride: 137 + 2 * 35.5 = 137 + 71 = 208.
2) Find the amount of barium chloride substance. To do this, divide its mass by its molar mass: 2.08 / 208 = 0.01 mol.
3) According to the reaction equation, there is one mole of barium sulfate per mole of barium chloride. This means that the amount of barium sulfate substance is 0.01 mol.
4) Molar mass of barium sulfate: 137 + 32 + 4 * 16 = 233.
5) Sludge mass: 233 * 0.01 = 2.33g – answer.