A 25.0-mL sample of sodium sulfate solution was analyzed by adding an excess of barium chloride solution
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Na2SO4(aq) + BaCl2(aq) → 2NaCl(aq) + BaSO4(s)
If 5.719 g of barium sulfate was obtained, what was the molarity of the original Na2SO4 solution?
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Convert the 5719 g of BaSO 4 to moles of BaSO 4 which is ...View the full answer
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