Đáp án:
C
Giải thích các bước giải:
\(\begin{array}{l}
F{e_2}{O_3}(a\,mol),F{e_3}{O_4}(b\,mol)\\
4Fe + 3{O_2} \to 2F{e_2}{O_3}\\
2a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\\
3Fe + 2{O_2} \to F{e_3}{O_4}\\
3b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b\\
(2a + 3b) \times 56 = 28\\
160a + 232b = 39,2\\
= > a = b = 0,1\\
\% mF{e_2}{O_3} = \dfrac{{0,1 \times 160}}{{39,2}} \times 100\% = 40\% \\
\% mF{e_3}{O_4} = 100 - 40 = 60\%
\end{array}\)