$ n_{CH_3COOH} = \dfrac{m}{M} =\dfrac{6}{60} = 0,1 (mol) $
$ CH_3COOH + C_2H_5OH \buildrel{{H_2SO_4d,t^o}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O $
Theo phương trình : $ n_{CH_3COOC_2H_5} = n_{CH_3COOH} = 0,1 (mol) $
$ \rightarrow m_{ltCH_3COOC_2H_5} = n × M = 0,1 × 88 = 8,8 (g) $
$ \rightarrow H = \dfrac{m_{tt}}{m_{lt}}×100\% = \dfrac{4,4}{8,8} × 100 = 50\% $