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Preparation of Sulphur Dioxide

An aqueous so...

Question

An aqueous solution of a gas $(X)$ gives the following reactions :
(i) It turns an acidified $K_{2} C r_{2} O_{7}$ solution green.
(ii) On boiling it with $H_{2} O_{2}$ , cooling it and then adding an aqueous solution of $B a C l_{2}$ , a precipitate insoluble in dilute hydrochloric acid is obtained.
(iii) On passing $H_{2} S$ in the solution a white turbidity is obtained.
Identify (X).

S O_{2}

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C O_{2}

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S i O_{2}

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None of the above

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Solution

The correct option is A $S O_{2}$
The compound (X) is $S O_{2}$ gas.

(i) It reduces Cr(VII) to Cr(III) and hence, the solution turns green.
3 $S O_{2} + C r_{2} O_{7}^{2 -} + 2 H^{+} ⟶ 2 C r^{3 +} + H_{2} O + 3 S O_{4}^{2 -}$

(ii) On boiling sulphur dioxide with hydrogen peroxide, sulphate ions are formed which combines with barium ions to form white precipitate of barium sulphate.

$S O_{2} + H_{2} O_{2} ⟶ S O_{4}^{2 -} + 2 H^{+}$

$S O_{4}^{2 -} + B a^{2 +} ⟶ B a S O_{4} ↓$

(iii) Sulphur dioxide gas reacts with hydrogen sulphide to form white turbidity of S.
$2 H_{2} S + S O_{2} ⟶ 3 S + 2 H_{2} O$

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Similar questions

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H_{2} S O_{4}

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