posted by sara
Aqueous sodium sulfate reacts with aqueous barium chloride to produce solid barium sulfate and aqueous sodium chloride. the solution contains 3.44g of barium chloride. you have a 0.282M solution of sodium sulfate. determine the volume of the sodium sulfate solution necessary to react with all of the barium chloride
BaCl2 + Na2SO4 => 2NaCl + BaSO4
Moles BaCl2 = Moles Na2SO4
Moles = Molarity x Volume(liters)
Moles BaCl2 = (3.44gms/207g/mol)= 0.0166 mole BaCl2 which will use the same number of moles Na2SO4.
Therefore, Moles Na2SO4 used = 0.0166 mole = (0.282M Na2SO4)(Volume in Liters) => Vol(L) = 0.059L = 59-ml of the 0.282M Na2SO4 solution.