$nCO2=0,11(mol$)$CO2+NaOH\rightarrow NaHCO3$
$x$ $x$ (mol)
$CO2+2NaOH\rightarrow Na2CO3+H2O$
$y$ $y$ (mol)
Theo bài ra ta có:$\begin{cases}x+y=0,11 \\ 84x+106y=11,44 \end{cases}\Leftrightarrow \begin{cases}x=0,01 \\ y=0,1\end{cases}$
$\Rightarrow \begin{cases}m NaHCO3=0,01.84=0,84(g) \\ m Na2CO3=0,1.106=10,6(g)\end{cases}$