Tuesday
March 28, 2017

Post a New Question

Posted by on .

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)

  • Chemistry - ,

    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

  • Chemistry - ,

    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

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question