A mixture of and is dissolved in water. Excess solution is added to precipitate all sulphate as . The precipitate is filtered off. The filtrate is evaporated to dryness. The residue weighs more th

Question

A mixture of  and  is dissolved in water. Excess  solution is added to precipitate all sulphate as . The precipitate is filtered off. The filtrate is evaporated to dryness. The residue weighs  more than the original mixture. What was the percentage of  in the original mixture if the mixture weighed ?

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Strategy:
 In this problem, one salt () is chemically converted into another () in the filtrate, while a precipitate () is removed. The change in the total residue mass is the difference between the mass of new ions added and old ions removed from the solution.

Step 1: Analyze the ionic exchange
Reaction: .
In the filtrate, 1 mole of  (mass 96) is replaced by 2 moles of  (mass 71) after the Barium is removed.
Net change per mole of .

Step 2: Relate to the mass increase
Given the final residue

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