A solution of sodium sulfate was treated with 0.29M Barium Chloride solution until all the sulfate ion had reacted. The reaction required 32.16mL of the Barium Chloride solution. How many grams of Sodium sulfate were in the solution?
BaCl2 + Na2SO4 ==> BaSO4 + 2NaCl
millimoles BaCl2 = mL x M = 0.29 x 32.16 = 16
You have 1 mol BaSO4 produced for every mols BaCl2 used; therefore, you must have produced 16 millimols (0.016 mols) BaSO4.
grams BaSO4 = mols BaSO4 x molar mass BaSO4 = ?