When $S O_{2}$ is passed through acidified solution of $H_{2} S$ :

H_{2} S O_{3}

is formed

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H_{2} S O_{4}

is formed

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sulphur is precipitated

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H_{2} S O_{5}

is formed

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Solution

The correct option is C sulphur is precipitated
$2 H_{2} S + S O_{2} ⟶ 3 S ↓ + 2 H_{2} O$
When $S O_{2}$ is passed through acidified solution of $H_{2} S$ , sulphur is precipitated.

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