nCO2 = 0,11 (mol)
CO2 + NaOH \(\rightarrow\)NaHCO3
x x (mol)
CO2 + 2NaOH\(\rightarrow\) Na2CO3 + H2O
y y (mol)
Theo đề ta có :
\(\left\{{}\begin{matrix}x+y=0,11\\84x+106y=11,4\end{matrix}\right.\)\(\left\{{}\begin{matrix}x=0,11\\y=0,1\end{matrix}\right.\)
=>mNaHCO3 = 0,01 . 84 = 0,84 (g)
=> mNa2CO3 = 0,1 . 106 = 10,6 (g)
=>% mNaHCO3 =\(\frac{0,84}{11,44}.100=7,34\%\)
=> %mNa2CO3 =100%-7,34%=93,66%