n CH3COOH = a(mol) ; n C2H5OH = b(mol)
=> 60a + 46b = 21,2(1)
$2CH_3COOH + 2Na \to 2CH_3COONa + H_2$
$2C_2H_5OH + 2Na \to 2C_2H_5ONa + H_2$
Theo PTHH :
n H2 = 0,5a + 0,5b = 4,48/22,4 = 0,2(2)
Từ (1)(2) suy ra a = b = 0,2
m CH3COOH = 0,2.60 = 12(gam)
m C2H5OH = 0,2.46 = 9,2(gam)