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Bond Pairs and Lone Pairs

10-3 mole mix...

Question

$10^{- 3}$ mole mixture of ozonized oxygen is used for oxidation of $I_{3}^{-}$ to $I_{2}$ which on reacting with 0.1 M hypo solution required 48 ml for complete reaction. The mole % of $O_{3}$ in original mixture is -
[Assume $O_{2}$ does not interfere in the reaction and Given : $2 e^{-} + O_{3} \to O_{2} + O^{2 -}]$

A

80%

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B

20%

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C

60%

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D

40%

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Solution

The correct option is A 80%
Reduction:
$2 e^{-} + O_{3} \to O_{2} + O^{2 -}$
oxidation:
$2 I_{3}^{-} \to 3 I_{2} + 2 e^{-}$
Net reaction
$2 I_{3}^{-} + O_{3} \to 3 I_{2} + O_{2} + O^{2 -}$

moles of hypo required for $I_{2}$ generated = $0.048 \times 0.1$
Again,
$I_{2} + 2 N a_{2} S_{2} O_{3} \to N a_{2} S_{4} O_{6} + 2 N a I$
So, moles of $I_{2}$ formed after reaction = 0.0024
since 1 mole of $I_{2}$ reacts with 2 mole of hypo.

From stoichiometry, 3 moles $I_{2}$ is formed from 1 mole of $O_{3}$ .
Hence moles of $O_{3}$ in the original mixture = $\frac{1}{3} \times 0.0024$ = 0.0008

percentage of $O_{3}$ in the original mixture = $\frac{0.008}{10^{- 3}} \times 100 = 80 %$

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