Question

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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 1

Answer:

dear mate here is your answer

Explanation:

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Answer 2

Answer:

Precipitate = Fe(OH)³ + BaSO⁴

Mass : Fe(OH)³ - 0.535g

BaSO⁴ - 1.748g

Explanation:

The precipitate is ferric hyroxide and barium sulphate