Question

On heating 4.9 gms of $KCl{O}_3$, the weight loss is found to be 0.384 gms. The $\%$ of $KCl{O}_3$ decomposed is ___

Answer 1

$2KCl{O}_3\to 2KCl+3O_2$
The reduction in weight is due the escape of Oxygen from the container.
Hence the weight of $O_2$ is 0.384 gm
moles of $O_2=\frac{wt}{molwt}=\frac{0.384}{32}=0.012moles$
2 moles of $KCl{O}_3$ gives 3 moles of $O_2$.
x moles of $KCl{O}_3$ gives 0.012 moles
$\frac{2}{x}=\frac{3}{0.012}$
$x=0.012\times \frac{2}{3}$
x = 0.008 moles
Wt of $KCl{O}_3$ decomposed = mol wt $\times$ moles = 122.5 $\times$ 0.008 = 0.98 gm
$\%$ dissociation = $\frac{decomposedwt}{totalwt}=\frac{0.98}{4.9}\times 100=20\%$.