Em tham khảo nha :
\(\begin{array}{l}
a)\\
FeC{l_2} + 2KOH \to Fe{(OH)_2} + 2KCl\\
{n_{FeC{l_2}}} = \dfrac{{25,4}}{{127}} = 0,2mol\\
{n_{KOH}} = \dfrac{{28}}{{56}} = 0,5mol\\
\dfrac{{0,2}}{1} < \dfrac{{0,5}}{2} \Rightarrow KOH\text{ dư}\\
{n_{Fe{{(OH)}_2}}} = {n_{FeC{l_2}}} = 0,2mol\\
{m_{Fe{{(OH)}_2}}} = 0,2 \times 90 = 18g\\
b)\\
TH1:\\
4Fe{(OH)_2} + {O_2} + 2{H_2}O \to 4Fe{(OH)_3}\\
2Fe{(OH)_3} \to F{e_2}{O_3} + 3{H_2}O\\
{n_{Fe{{(OH)}_3}}} = {n_{Fe{{(OH)}_2}}} = 0,2mol\\
{n_{F{e_2}{O_3}}} = \frac{{{n_{Fe{{(OH)}_3}}}}}{2} = 0,1mol\\
{m_{F{e_2}{O_3}}} = 0,1 \times 160 = 16g\\
TH2:\\
Fe{(OH)_2} \to FeO + {H_2}O\\
{n_{FeO}} = {n_{Fe{{(OH)}_2}}} = 0,2mol\\
{m_{FeO}} = 0,2 \times 72 = 14,4g
\end{array}\)
