PTHH: 2NaOH + CO2 \(\rightarrow\) Na2CO3 + H2O (1)
NaOH + CO2 \(\rightarrow\) NaHCO3 (2)
nCO2=\(\frac{4,48}{22,4}=0,2\left(mol\right)\)
Đặt nNa2CO3=a
nNaHCO3=b
Ta có: \(\left\{{}\begin{matrix}a+b=0,2\\106a+84b=17,9\end{matrix}\right.\)
Giải hệ ta được : \(\left\{{}\begin{matrix}a=0,05\\b=0,15\end{matrix}\right.\)
Theo PT (1): nNaOH(1)=2nNa2CO3=0,1(mol)
Theo PT (2):nNaOH(2)=nNaHCO3=0,15(mol)
\(\Rightarrow n_{NaOH}=0,1+0,15=0,25\left(mol\right)\)
\(\Rightarrow\)CM dd NaoH=\(\frac{0,25}{0,5}=0,5M\)