đây e nhé $CH_3COOH + C_2H_5OH \leftrightharpoons CH_3COOC_2H_5 + H_2O$
bd $1$ $1$
CB $1/3$ $1/3$ $2/3$ $2/3 $
$K_{cb} = \frac{2/3.2/3}{1/3.1/3} = 4$
$CH_3COOH + C_2H_5OH \leftrightharpoons CH_3COOC_2H_5 + H_2O$
bd $1$ $x$
CB $0,1$ $x-0,9$ $ 0,9$ $0,9$
$K_{cb} = \frac{0,9.0,9}{0,1(x-0,9} = 4 => x = 2,925$ mol