Cho $\begin{cases}NO:x(mol)\\ NO_2:y(mol)\\\end{cases}$
`n_{hh\ khí}=0,54(mol)`
`=> x+y=0,54(mol)(1)`
`∑_{hh\ khí}=\frac{30x+46y}{x+y}`
`=> 10,167.4=\frac{30x+46y}{x+y}`
`=> 21,96=30x+46y(2)`
`(1),(2)=>` $\begin{cases}x=0,18(mol)\\y=0,36(mol)\\\end{cases}$
Quy đổi hỗn hợp `A` về : $\begin{cases}Fe:a(mol)\\O:b(mol)\\\end{cases}$
`=> 56a+16b=104,8g(3)`
BTe:
$\mathop{Fe}\limits^{0}\to \mathop{Fe}\limits^{+3}+3e$
$\mathop{N}\limits^{+5}+3e\to \mathop{N}\limits^{+2}$
$\mathop{N}\limits^{+5}+e\to \mathop{N}\limits^{+4}$
$\mathop{O}+2e\to \mathop{O}\limits^{-2}$
`=> 2b+3n_{NO}+n_{NO_2}=3a`
`=>3a-2b=0,9(mol)(4)`
`(3),(4)=> `$\begin{cases}a=1,4(mol)\\ b=1,65(mol)\\\end{cases}$
`=> x= 56a`
`=> x=56.1,4=78,4g`
