Đáp án:

\(\begin{array}{l}
\% {m_{FeO}} = 31,03\% \\
\% {m_{F{e_2}{O_3}}} = 68,97\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
hh:FeO(a\,mol),F{e_2}{O_3}(b\,mol)\\
{m_{hh}} = 46,4g \Rightarrow 72a + 160b = 46,4(1)\\
T{N_1}:\\
FeO + {H_2} \xrightarrow{t^0} Fe + {H_2}O\\
F{e_2}{O_3} + {H_2} \xrightarrow{t^0} 2Fe + 3{H_2}O\\
{n_{Fe}} = \dfrac{{16,8}}{{56}} = 0,3\,mol\\
 \Rightarrow 0,5a + 0,5 \times 2 \times b = 0,3 \Rightarrow 0,5a + b = 0,3(2)\\
T{N_2}:\\
4FeO + {O_2} \xrightarrow{t^0} 2F{e_2}{O_3}\\
{n_{F{e_2}{O_3}}} = \dfrac{{24}}{{160}} = 0,15\,mol\\
 \Rightarrow 0,5a \times \dfrac{1}{2} + 0,5b = 0,15 \Rightarrow 0,5a + b = 0,3(3)\\
(1),(2) \Rightarrow a = b = 0,2\\
\% {m_{FeO}} = \dfrac{{0,2 \times 72}}{{46,4}} \times 100\%  = 31,03\% \\
\% {m_{F{e_2}{O_3}}} = 100 - 31,03 = 68,97\% 
\end{array}\)