On heating $4.9 g$ of ${KClO}_3$, it shows a weight loss of $0.384 g$. What percentage of ${KClO}_3$ has decomposed?

The balanced reaction is as follows:

$2{KClO}_3 \left( s\right)\to 2KCl\left(s\right) +3O_2\left( g\right)$

From the above reaction we can say that $2 \mathrm{mol} {\mathrm{KClO}}_3 \mathrm{will} \mathrm{produce} 2 \mathrm{mol} \mathrm{KCl} \mathrm{and} 3 {\mathrm{molO}}_2$

Using the formula: $\mathrm{moles}=\frac{\mathrm{given} \mathrm{mass}}{\mathrm{molecular} \mathrm{mass}}$

Therefore moles of oxygen produced = $\frac{0.384}{32}=0.012mol$

From the reaction we can calculate the decomposed moles of ${KClO}_3$ which is $0.012\times \frac{2}{3}=0.008mol$

Now we calculate mass of ${KClO}_3$ = moles $\times$ molecular mass= 0.008$\times$123=0.980gm

Therefore percentage of ${KClO}_3$ decomposed = $\frac{0.980}{4.9}\times 100=20\%$

Hence the correct option is (D).