$4 g$ of a mixture of ${Na}_2{CO}_3$ and ${NaHCO}_3$ on heating liberates $448ml$ of ${CO}_2$ at $STP$. The percentage of ${Na}_2{CO}_3$ in the mixture is

$moles=\frac{given volume}{22400\left(ml\right)}$

Therefore moles= $\frac{448}{22400}=0.02$

$\mathrm{Decomposition} \mathrm{reaction} \mathrm{of} {\mathrm{NaHCO}}_3 \mathrm{is} \mathrm{as} \mathrm{follows}: 2\mathrm{NaHCO}{}_3\to {\mathrm{Na}}_2{\mathrm{CO}}_3+{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}$

From the above reaction, 1 mole ${\mathrm{CO}}_2 \mathrm{is} \mathrm{equivalent} \mathrm{to} 2$ moles ${NaHCO}_3$

Therefore, $0.02$ mole ${\mathrm{CO}}_2 \mathrm{is} \mathrm{equivalent} \mathrm{to} 0.04$ mole ${NaHCO}_3$

Now using the formula : $moles=\frac{given mass}{molecular mass}$

Mass of ${NaHCO}_3=0.04\times 84=3.36gm$

Mass of ${Na}_2{CO}_3=4-3.36=0.64gm$

Therefore percentage of ${Na}_2{CO}_3$ in the mixture is $\frac{0.64}{4}\times 100=16\%$

Hence the correct option is (B).