với CTHH: FeO
\(\%m_{Fe}=\dfrac{56\cdot100\%}{56+16}=77,77\%\)
với CTHH: Fe2O3
\(\%m_{Fe}=\dfrac{56\cdot2\cdot100\%}{56\cdot2+16\cdot3}=70\%\)
=> D
\(FeO\)
\(\text{%}Fe=\dfrac{56.1}{72}.100\text{%}=77,77\text{%}\)
\(Fe_2O_3\)
\(\text{%}Fe=\dfrac{2.56}{160}.100\text{%}=70\text{%}\)
\(\Rightarrow D\)