Em tham khảo nha :
\(\begin{array}{l}
1)\\
F{e_x}{O_y} + y{H_2} \to xFe + y{H_2}O\\
{n_{{H_2}O}} = \dfrac{{14,4}}{{18}} = 0,8mol\\
{n_{F{e_x}{O_y}}} = \dfrac{{{n_{{H_2}O}}}}{y} = \dfrac{{0,8}}{y}mol\\
{M_{F{e_x}{O_y}}} = 46,4:\dfrac{{0,8}}{y} = 58y\,dvC\\
\Rightarrow 56x + 16y = 58y\\
x = 3 \Rightarrow y = \dfrac{{56 \times 3}}{{58 - 16}} = 4\\
CTHH:F{e_3}{O_4}\\
2)\\
F{e_x}{O_y} + yCO \to xFe + yC{O_2}\\
{n_{Fe}} = \dfrac{{89,6}}{{56}} = 1,6mol\\
{n_{F{e_x}{O_y}}} = \dfrac{{{n_{Fe}}}}{x} = \dfrac{{1,6}}{x}mol\\
{M_{F{e_x}{O_y}}} = 128:\dfrac{{1,6}}{x} = 80x\\
56x + 16y = 80x\\
x = 2 \Rightarrow y = \dfrac{{(80 - 56) \times 2}}{{16}} = 3\\
CTHH:F{e_2}{O_3}
\end{array}\)