$Na_2CO_3+2HCl\rightarrow 2NaCl+CO_2\uparrow+H_2O$ $x 2x 2x x$
$\Rightarrow \%NaCl=\frac{2x.58,5}{200+120-44x}=\frac{20}{100}\Rightarrow x=0,5 mol$
$\Rightarrow \begin{cases}\%Na_2CO_3=\frac{0,5.106.100}{200}=26,5\% \\ \%HCl=\frac{2.0,5.36,5.100}{120}=30,42\% \end{cases}$