\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(m_{dd.NaOH}=50.1,28=64\left(g\right)\)
=> \(n_{NaOH}=\dfrac{64.10\%}{40}=0,16\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,16}{0,1}=1,6\)
=> Tạo ra muối Na2CO3 và NaHCO3
PTHH: 2NaOH + CO2 --> Na2CO3 + H2O
0,16--->0,08---->0,08
Na2CO3 + CO2 + H2O --> 2NaHCO3
0,02<---0,02------------->0,04
=> \(\left\{{}\begin{matrix}m_{Na_2CO_3}=\left(0,08-0,02\right).106=6,36\left(g\right)\\m_{NaHCO_3}=0,04.84=3,36\left(g\right)\end{matrix}\right.\)
mdd sau pư = 0,1.44 + 64 = 68,4 (g)
\(\left\{{}\begin{matrix}C\%_{Na_2CO_3}=\dfrac{6,36}{68,4}.100\%=9,3\%\\C\%_{NaHCO_3}=\dfrac{3,36}{68,4}.100\%=4,9\%\end{matrix}\right.\)
