Chemistry

posted by Beth

A solution contains an unknown mass of dissolved barium ions. When sodium sulfate is added to the solution, a white precipitate forms. The precipitate is filtered and dried and found to have a mass of 258 mg. What mass of baium was in the original solution? (Assume that all of the barium was precipitated out of solution by the reaction)

  1. DrBob222

    258 mg BaSO4 = 0.258 g.
    mols BaSO4 = 0.258/molar mass BaSO4.
    mols Ba = same as mols BaSO4,
    g Ba = mols x atomic mass Ba

  2. Paris

    Na2SO4(aq) + Ba(aq)-> BaSO4(s) + 2Na(aq)

    .258gr BaSO4 x 1 mol BaSO4/ 233.4 gr BaSO4 x 1 mol Ba/1 mol BaSO4 x 137.33 grams Ba/1 mol Ba = .152 grams of Ba

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