Đáp án:
a, \(\begin{array}{l}
\dfrac{{{m_{Fe}}}}{{{m_O}}} = \dfrac{{21}}{8} \to \dfrac{{{n_{Fe}}}}{{{n_O}}} = \dfrac{{21}}{8}:\dfrac{{56}}{{16}} = \dfrac{3}{4}\\
F{e_3}{O_4}
\end{array}\)
b,
\(\begin{array}{l}
\% P = \dfrac{{31 \times {n_P}}}{{142}} \times 100\% = 43,66\% \to {n_P} = 2\\
\% O = \dfrac{{16 \times {n_O}}}{{142}} \times 100\% = 56,34\% \to {n_O} = 5\\
{P_2}{O_5}
\end{array}\)