\(n_{NaOH}=0.25\cdot2.4=0.6\left(mol\right)\)
\(NaOH+CO_2\rightarrow NaHCO_3\)
Hấp thụ tối thiểu => tỉ lệ 1 : 1
\(n_{CO_2}=0.6\left(mol\right)\)
\(\Rightarrow n_{tb}=\dfrac{1}{2}\cdot0.6=\dfrac{0.3}{n}\left(mol\right)\)
\(m_{tb}=\dfrac{0.3}{n}\cdot162n=48.6\left(g\right)\)
\(H\%=\dfrac{48.6}{60}\cdot100\%=81\%\)