Acidified $K_{2} C r_{2} O_{7}$ oxidizes $H_{2} S$ to

S O_{2}

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S O_{3}

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S

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H_{2} S O_{4} + S

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Solution

The correct option is C $S$
$3 H_{2} S + C r_{2} O_{7}^{2 -} + 8 H^{+} \to 2 C r^{3 +} + 3 S + 7 H_{2} O$

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