gọi `x,y` là số mol `C_2H_5OH,CH_3COOH`

`=>46x+60y=54,4(1)`

`PTHH:2C_2H_5OH+2Na->2C_2H_5ONa+H_2`

`(mol)`  `x->`                                                                `0,5x`

          `2CH_3COOH+2Na->2CH_3COONa+H_2`

`(mol)`     `y->`                                                                        `0,5y`

`n_{H_2}={11,2}/{22,4}=0,5(mol)`

`=>0,5x+0,5y=0,5(2)`

`(1)(2)=>x=0,4.y=0,6`

`%m_{C_2H_5OH}={0,4.46.100%}/{54,4} ~~33,82%`

`%m_{CH_3COOH}=100-33,82=66,18%`

`b.`

$\rm PTHH:C_2H_5OH+CH_3COOH \buildrel{{H_2SO_4 đ,t^o}}\over\rightleftharpoons CH_3COOC_2H_5+H_2O$

ta có: `{0,4}/1 <{0,6}/1 =>CH_3COOH` dư

`n_{CH_3COOC_2H_5}=n_{C_2H_5OH}=0,4(mol)`

`m_{CH_3COOC_2H_5} l t=0,4.88=35,2(g)`

`H=85%->m_{CH_3COOC_2_5}=35,2.85%=29,92(g)`