$n_{C_2H_5OH} = n_{CH_3COOH} = 2n_{H_2} = 0,5$ mol => $m_{C_2H_5OH} = \frac{0,5.46.100}{25} = 92$ gam
Gọi $n_{C_2H_5OH} =
a$ mol $C_2H_5OH + O_2 => CH_3COOH + H_2O$bd $a$CB $0,75a 0,25a 0,25a$$C_2H_
5OH,CH_3COOH
,H_2O+Na$ tạo $C_2H_5ONa,CH_3COONa,NaOH+1/2H_2$$n_Y = 0,75a+0,25a+0,25a = 2n_{H_2} = 0,
25$ mol => $
a = 0,2
$ mol=> $m = 9
,2$ gam