Sửa `HN->HNO_3`
`n_{Fe}=\frac{28}{56}=0,5(mol)`
$X : \ \begin{cases}NO: \ x(mol)\\N_2: \ y (mol)\\\end{cases}$
`n_{X}=\frac{2,24}{22,4}=0,1(mol)`
`->x+y=0,1`
`M_{X}=1.29=29(g//mol)`
Theo sơ đồ đường chéo
`->\frac{x}{y}=\frac{29-28}{30-29}=1`
`->x=y=0,05(mol)`
$Fe^{+0}\to Fe^{+3}+3e$
$O^{0}+2e \to O^{-2}$
$N^{+5}+3e\to N^{+2}$
$2N^{+5}+10e\to 2N^{+0}$
BTe
`->3n_{Fe}=2n_{O}+3n_{NO}+10n_{N_2}`
`->n_{O}=\frac{3.0,5-3.0,05-10.0,05}{2}=0,425(mol)`
`->m_{O}=6,8(g)`
`->m=28+6,8=34,8(g)`