R2On+nCO\(\rightarrow\)2R+nCO2
nO trong R2On=\(\frac{\text{24-16,8}}{16}\)=0,45(mol)
\(\rightarrow\)nR2On\(=\frac{0,45}{n}\)(mol)
MR2On=24:\(\frac{0,45}{n}\)=\(\frac{160n}{3}\)
Ta có 2R+16n=\(\frac{160n}{3}\)
\(\rightarrow\)R=\(\frac{56n}{3}\)
n=3 \(\rightarrow\) R=56
\(\Rightarrow\) R là Fe
Vậy CT oxit là Fe2O3