nNO = 2.24/22.4 = 0.1 (mol)
N+5 + 3e ---> N+2
0.1 --- 0.3 --- 0.1 (mol)
Ta có: \(\left\{{}\begin{matrix}56x+16y=21.6\\3x-2y=0.1\times3\left(BTe\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0.3\\y=0.3\end{matrix}\right.\rightarrow x:y=1:1\rightarrow CT:FeO\)