$Na + H_{2}O \rightarrow NaOH + \frac{1}{2} H_{2}$
$Ba + 2H_{2}O \rightarrow Ba(OH)_{2} + H2$
$n_{OH^{-}} = 0.1 + 0.2 \times 2 = 0.5 mol$
$Fe^{2-} + 2OH^{-} \rightarrow Fe(OH)_{2} $
$\Rightarrow n_{FeCl_{2}} pư= n_{Fe(OH)_{2}} =\frac{0.5}{2}= 0.25 mol$
$m = m_{Fe(OH)_{2}} = 22.5 g$
C