Em tham khảo nha :
\(\begin{array}{l}
{n_{C{O_2}}} = \dfrac{{2,688}}{{22,4}} = 0,12mol\\
C{O_2} + 2NaOH \to N{a_2}C{O_3} + {H_2}O\\
NaOH + C{O_2} \to NaHC{O_3}\\
hh:N{a_2}C{O_3}(a\,mol);NaHC{O_3}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,12\\
106a + 84b = 11,84
\end{array} \right.\\
\Rightarrow a = 0,08;b = 0,04\\
{n_{NaOH}} = 2{n_{N{a_2}C{O_3}}} + {n_{NaHC{O_3}}} = 0,2mol\\
{V_{NaOH}} = \dfrac{{0,2}}{1,1} = 0,1818l = 181,8ml\\
{C_{{M_{N{a_2}C{O_3}}}}} = \dfrac{{0,08}}{{0,1818}} = 0,44M\\
{C_{{M_{NaHC{O_3}}}}} = \dfrac{{0,04}}{{0,1818}} = 0,22M
\end{array}\)