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

100% of Mixture = 4g

$2{\mathrm{NaHCO}}_3\to {\mathrm{Na}}_2{\mathrm{CO}}_3+{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2$

$2\times 84 \mathrm{g}\to 44\mathrm{g} \mathrm{of} {\mathrm{CO}}_2$

$2\times 84 \mathrm{g}\to 22400\mathrm{mL} \mathrm{of} {\mathrm{CO}}_2$

$?\to 448 \mathrm{mL} \mathrm{of} {\mathrm{CO}}_2$

$\mathrm{x}=\frac{2\times 84\times 448}{22400}=3.36\mathrm{g} \mathrm{of} {\mathrm{NaHCO}}_3$

Weight of ${\mathrm{Na}}_2{\mathrm{CO}}_3=4-3.36=0.64\mathrm{g}$

100% → 4 g

$16\% =\frac{0.64\times 100}{4}=?\to 0.64\mathrm{g}$