Question

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)

Answer 1

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

Answer 2

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 3

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 then found to have a mass of 266 mg.

Answer: m(Ba+2)=157 mg

Balanced Chemical Equation: Ba + Na2SO4 --> BaSO4 + 2 Na