Question

$2KCl{O}_3\stackrel{\Delta }{\to }2KCl+3O_2;C+O_2\stackrel{\Delta }{\to }C{O}_2$.

Calculate the amount of $KCl{O}_3$ 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]

• A. 326.67 g
• B. 81.62 g
• C. 98.34 g
• D. 163.33 g

Answer 1

The correct option is D 163.33 g

Reaction:
$2KCl{O}_3\stackrel{\Delta }{\to }2KCl+3O_2$
2 2 3
$C+O_2\stackrel{\Delta }{\to }C{O}_2$
1 1
$24$ gm = $2$ moles of carbon
so $2$ mole $C$ requires $2$ mole $O_2$ and this requires $KCl{O}_3=\frac{4}{3}mole$ = $4\times \frac{122.5}{3}$ = $163.3gm$