CH3COOH + Na CO3 -> CH3COONa + H2O + Co2\(\uparrow\)
1 1 1 1 1 (mol)
o.1 0,1 0,1 0,1 0,1 (mol)
Ta có : n Co2 = V/22,4 = 2,34/22,4 \(\approx0.1\left(mol\right)\)
=> m CH3COOH =n.M \(\approx\)0,1 60 = 6( g)
=> m% CH3COOH \(=\dfrac{6.100}{30,4}\approx19,74\%\)
=> m% C2H5OH \(\approx\)100% -19.74%=80,26%