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A solution of pure ferric sulphates containing 0.280 g of ferric ions is treated with an excess of a solution of Ba(OH)2. Name the precipitate formed and calculate its weight.
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Answer:
Precipitate = Fe(OH)³ + BaSO⁴
Mass : Fe(OH)³ - 0.535g
BaSO⁴ - 1.748g
Explanation:
The precipitate is ferric hyroxide and barium sulphate
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