a) \(n_{CO_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
\(n_{NaOH}=2.0,2=0,4\left(mol\right)\)
Xét tỉ lệ \(\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,4}{0,6}=0,67\) => Tạo ra muối NaHCO3
b)
PTHH: NaOH + CO2 --> NaHCO3
0,4------------->0,4
=> \(m_{NaHCO_3}=0,4.84=33,6\left(g\right)\)