`CH_3COOH + C_2H_5OH` $\buildrel{{H_2SO_4 đặc,t^o}}\over\rightleftharpoons$ `CH_3COOC_2H_5 + H_2O`

`n_{CH_3COOH} = (6)/(60) = 0,1` mol

Theo PTHH

`n_{CH_3COOC_2H_5(text[lí thuyết])} = n_{CH_3COOH} = 0,1` mol

`m_{CH_3COOC_2H_5(lí thuyết)} = 0,1 . 88 = 8,8g`

`=> H% = (6,6)/(8,8) . 100% = 75%`