$2 K C l O_{3} Δ \to 2 K C l + 3 O_{2}; C + O_{2} Δ \to C O_{2}$ .
Calculate the amount of $K C l O_{3}$ which on thermal decomposition gives 'X' volume of $O_{2}$ , which is the volume required for combustion of 24 g of carbon.
[K=39, Cl=35.5, O=16, C=12]

326.67 g

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81.62 g

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98.34 g

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163.33 g

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Solution

The correct option is D 163.33 g
Reaction:
$2 K C l O_{3} Δ \to 2 K C l + 3 O_{2}$
2 2 3
$C + O_{2} Δ \to C O_{2}$
1 1
$24$ gm = $2$ moles of carbon
so $2$ mole $C$ requires $2$ mole $O_{2}$ and this requires $K C l O_{3} = \frac{4}{3} m o l e$ = $4 \times \frac{122.5}{3}$ = $163.3 g m$

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