$n_{CO_2}=\dfrac{2,24}{22,4}=0,1(mol)$

$n_{NaOH}=\dfrac{200.3\%}{40}=0,15(mol)$

$\dfrac{n_{NaOH}}{n_{CO_2}}=1,5$

$\to$ phản ứng tạo 2 muối: $NaHCO_3$ ($x$ mol) và $Na_2CO_3$ ($y$ mol)

Bảo toàn $Na$: $x+2y=0,15$

Bảo toàn $C$: $x+y=0,1$

Giải hệ: $x=y=0,05$

$m_{\text{dd spứ}}=0,1.44+200=204,4g$

$C\%_{NaHCO_3}=\dfrac{0,05.84.100}{204,4}=2,05\%$

$C\%_{Na_2CO_3}=\dfrac{0,05.106.100}{204,4}=2,59\%$

Em tham khảo nha :

\(\begin{array}{l}
{m_{NaOH}} = \dfrac{{3 \times 200}}{{100}} = 6g\\
{n_{NaOH}} = \dfrac{6}{{40}} = 0,15mol\\
{n_{C{O_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
T = \dfrac{{{n_{NaOH}}}}{{{n_{C{O_2}}}}} = \dfrac{{0,15}}{{0,1}} = 1,5\\
 \Rightarrow\text{Phản ứng tạo ra 2 muối } NaHC{O_3},N{a_2}C{O_3}\\
NaOH + C{O_2} \to NaHC{O_3}\\
2NaOH + C{O_2} \to N{a_2}C{O_3} + {H_2}O\\
hh:NaHC{O_3}(a\,mol),N{a_2}C{O_3}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,1\\
a + 2b = 0,15
\end{array} \right.\\
 \Rightarrow a = 0,05;b = 0,05\\
C{\% _{NaHC{O_3}}} = \dfrac{{0,05 \times 84}}{{0,1 \times 44 + 200}} \times 100\%  = 2,05\% \\
C{\% _{N{a_2}C{O_3}}} = \dfrac{{0,05 \times 106}}{{0,1 \times 44 + 200}} \times 100\%  = 2,59\% \\

\end{array}\)