how much of a 1.50M solution sodium sulfate solution in milliliters is required to completely precipitate all the barium in 150.0mL of a 0.250 M barium nitrate solution

Answer 1

Ba(NO3)2 + Na2SO4 ==> BaSO4 + 2NaNO3

millimoles Ba(NO3)2 = mL x M = ?
mmoles Na2SO4 needed = the same since 1 mole Ba(NO3)2 = 1 mole Na2SO4.
mmoles Na2SO4 = mL x M
You know M, solve for mL.