A mixture of ${Na}_2{CO}_3$ and ${NaHCO}_3$ having a total weight of $100gm$ on heating produced 11.2 $L$ of ${CO}_2$ under STP conditions. The percentage of ${Na}_2{CO}_3$ in the mixture is

${Na}_2{CO}_3$ does not decompose on heating.

Therefore the decomposition reaction is as follows:

$2{NaHCO}_3⟶{Na}_2{CO}_3+H_{2}O+{CO}_2$

Using the formula: : $moles=\frac{given volume}{22.4l}$

Therefore moles=$\frac{11.2}{22.4}=0.5l$

From the above reaction we can say that $\mathrm{Two} \mathrm{moles} \mathrm{of} {\mathrm{NaHCO}}_3 \mathrm{decomposes} \mathrm{on} \mathrm{heating} \mathrm{to} \mathrm{give} \mathrm{one} \mathrm{moles} \mathrm{of} {\mathrm{CO}}_2.$

Therefore 0.5mol carbon dioxide gives 1mol of ${NaHCO}_3$

Therefore mass= 1$\times$84=84g

Mass of ${Na}_2{CO}_3$=100-84=16g

$\%N{a}_{2}C{O}_3=\frac{16}{100}\times 100=16\%$

Hence the correct option is (D).