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florrye florrye
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11 years ago
A 3.670 g sample of a mixture was analyzed for barium ion by adding a small excess of sulfuric acid to an aqueous solution of the sample. The resultant reaction produced a precipitate of barium sulfate, which was collected by filtration, washed, dried, and weighed.

If 0.3425 g of barium sulfate was obtained, what was the mass percentage of barium in the sample?
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julieyantachkjulieyantachk
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11 years ago
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11 years ago
Ba2+ (aq) + SO42- (aq) = BaSO4 (s)
moles BaSO4 = 0.3425 g / 233.43 g/mol=0.001467
= moles Ba2+ in the sample
mass Ba2+ = 0.001467 mol x 137.327 g/mol= 0.2015 g
% Ba2+ = 0.2015 x 100 / 3.670=5.490
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