pt 2CH3COOH+2Na\(\rightarrow\)2CH3COONa+H2
ta có nCH3COONa=\(\dfrac{32,8}{82}\)=0,4 mol
theo pt nCH3COOH=NCH3COONa =0,4 mol \(\Rightarrow\)Cm CH3COOH=\(\dfrac{0.4}{0,5}\)=0,8 mol/l
theo pt nH2 =1/2 n CH3COOH =0,2 MOL
\(\Rightarrow\)V H2 =0,2 *22,4=4,48 l