\(m_O=7,2-5,6=1,6\left(g\right)\)
\(n_{Fe}=\frac{5,6}{56}=0,1\left(mol\right)\)
\(n_O=\frac{1,6}{16}=0,1\left(mol\right)\)
\(n_{Fe}:n_O=0,1:0,1=1:1\)
CTHH:FeO
\(CTTQ:Fe_xO_y\)
Theo đề ta có:
\(\frac{56x}{56x+16y}=\frac{5,6}{7,2}\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
\(\rightarrow CTHH:FeO\)