A __ (black/white) precipitate of _(barium sulphate/ barium bisulphate) is formed when barium chloride is treated with sulphuric acid.

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Solution

When barium chloride ( ${BaCl}_{2}$ ) is treated with sulphuric acid ( $H_{2} {SO}_{4}$ ), a white precipitate of barium sulphate ( ${BaSO}_{4}$ ) along with hydrochloric acid ( $HCl$ ) is formed.
${BaCl}_{2} (aq) + H_{2} {SO}_{4} (aq) \to {BaSO}_{4} (aq) + 2 HCl (l) Barium chloride + sulphuric acid \to Barium sulphate + hydrochloric acid$
So, A white precipitate of barium sulphate is formed when barium chloride is treated with sulphuric acid.

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A ______ (black/white) precipitate of ______ ___(barium sulphate/ barium bisulphate) is formed when barium chloride is treated with sulphuric acid.

A __ (black/white) precipitate of _(barium sulphate/ barium bisulphate) is formed when barium chloride is treated with sulphuric acid.