A mixture of barium sulfate and barium carbonate weighing 120 g was treated with an excess of hydrochloric acid.
| December 10, 2020 | education
A mixture of barium sulfate and barium carbonate weighing 120 g was treated with an excess of hydrochloric acid. At the same time, 11.2 liters of gas were released. Determine the percentage of salts in the original mixture (in percent by weight).
BaSO4 + 2HCl = BaCl2 + H2SO4
BaCO3 + 2HCl = BaCl2 + H2O + CO2
M (BaSO4) = 233 G \ mol
M (BaCO3) = 197
n (gas) = 11.2 / 22.4 = 0.5 mol is carbon dioxide
hence n (BaCO3) = 0.5 mol
m (BaCO3) = 0.5 * 197 = 98.5 g
mass fraction of carbonate = 98.5 / 120 = 0.82
answer: 82% (carbonate) and 18% (sulfate)
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