a)
$FeO + 2HCl \to FeCl_2 + H_2O$
$FeCl_2 + 2NaOH \to Fe(OH)_2 + 2NaCl$
b) $n_{FeO} = \dfrac{1,44}{72} = 0,02(mol)$
$n_{HCl} = 2n_{FeO} = 0,02.2 = 0,04(mol)$
$\Rightarrow C_{M_{HCl}} = \dfrac{0,04}{0,2} = 0,2M$
c) $n_{Fe(OH)_2} = n_{FeO} = 0,02(mol)$
$m_{Fe(OH)_2} = 0,02.98 = 1,96(gam)$