a) $n_{FeO} = \dfrac{14,4}{72} = 0,2(mol)$
$FeO + 2HCl \to FeCl_2 + H_2O$
$n_{HCl} = 2n_{FeO} = 0,4(mol)$
$m_{dd\ HCl} = \dfrac{0,4.36,5}{20\%} = 73(gam)$
b) $m_{dd\ sau\ pư} = 14,4 + 73 = 87,4(gam)$
$n_{FeCl_2} = n_{FeO} = 0,2(mol)$
$C\%_{FeCl_2} = \dfrac{0,2.127}{87,4}.100\% = 29,06\%$