Em tham khảo nha :
\(\begin{array}{l}
a)\\
F{e_x}{O_y} + 2yHCl \to xFeC{l_{\frac{{2y}}{x}}} + y{H_2}O\\
{n_{HCl}} = 0,4 \times 2 = 0,8mol\\
{n_{F{e_x}{O_y}}} = \dfrac{{{n_{HCl}}}}{{2y}} = \frac{{0,4}}{y}mol\\
{M_{F{e_x}{O_y}}} = 23,2:\frac{{0,4}}{y} = 58y\\
\Rightarrow 56x + 16y = 58y\\
x = 3 \Rightarrow y = 4\\
CTHH:F{e_3}{O_4}\\
b)\\
F{e_3}{O_4} + 8HCl \to FeC{l_2} + 2FeC{l_3} + 4{H_2}O\\
{n_{FeC{l_2}}} = {n_{F{e_3}{O_4}}} = 0,1mol\\
{n_{FeC{l_3}}} = 2{n_{F{e_3}{O_4}}} = 0,2mol\\
{C_{{M_{FeC{l_3}}}}} = \dfrac{{0,2}}{{0,4}} = 0,5M\\
{C_{{M_{FeC{l_2}}}}} = \dfrac{{0,1}}{{0,4}} = 0,25M
\end{array}\)