Chọn `n_{C_5H_{11}OH} = n_{CH_3COOH} = 1 (mol)`

`m_{ddC_5H_{11}OH} = (88)/(98\%) = (4400)/(49) (g)`

`m_{ddCH_3COOH} = (60)/(96\%) = 62,5 (g)`

`m_{H_2O} = (4400)/(49) - 88 + 62,5 - 60 = (421)/(98) (g)`

`n_{H_2O} = (421)/(98) = (421)/(1764) (mol)`

$C_5H_{11}OH + CH_3COOH \buildrel{{H_2SO_4 (đặc),t^o }}\over\rightleftharpoons CH_3COOC_5H_{11} + H_2O$

`0,6 -> 0,6 -> 0,6`

`K = (0,6 . (0,6 + (421)/(1764) ))/(0,4 . 0,4) = 3,145`

Thực tế:

`n_{C_5H_{11}OH} = (440.98\%)/(88) = 4,9(kmol)`

Đặt `n_{CH_3COOH} = a (kmol)`

`m_{dd} = (60a)/(96\%) = 62,5a (kg)`

`m_{H_2O} = 440-4,9..88 + 62,5a - 60a =8,8 + 2,5a (kg)`

`n_{H_2O} = ( 8,8 + 2,5a)/(18) = (22)/(45) + (5a)/(36) (kmol)`

`n_{C_5H_{11}OH(pư)} = 4,9.80\% = 3,92 (kmol)`

$C_5H_{11}OH + CH_3COOH \buildrel{{H_2SO_4 (đặc),t^o }}\over\rightleftharpoons CH_3COOC_5H_{11} + H_2O$

`3,92 -> 3,92 -> 3,92 -> 3,92`

`=> K = (3,92 . ( (5a)/(36) + (22)/(45) + 3,92))/( (4,9-3,92) . (a - 3,92) ) = 3,145`

`=> a = 11,5716`

`=> m = 11,5716.62,5 = 723,225 (g)`