$CH_3COOH + C_2H_5OH \buildrel{{H_2SO_4,t^o}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O$
$n_{CH_3COOC_2H_5} = \dfrac{6,6}{88} = 0,075(mol)$
$n_{CH_3COOH\ đã\ dùng} = n_{C_2H_5OH\ đã\ dùng} = 0,075 :75\% = 0,1(mol)$
$m_{CH_3COOH} = 0,1.60 = 6(gam)$
$m_{C_2H_5OH} = 0,1.46 =4 ,6(gam)$