$Fe+\dfrac{1}{2}O_2 \xrightarrow{t^o}FeO$

$3Fe+2O_2\xrightarrow{t^o}Fe_3O_4$

$4Fe+3O_2 \xrightarrow{t^o}2Fe_2O_3$

$FeO+4HNO_3 \to Fe(NO_3)_3+NO+2H_2O$

$3Fe_3O_4+28HNO_3 \to 9Fe(NO_3)_3+NO+14H_2O$

$Fe_2O_3+6HNO_3 \to 2Fe(NO_3)_3+3H_2O$

$n_{NO}=\dfrac{5.6}{22.4}=0.25(mol)$

Bảo toàn $e:$

$2n_{O}=3n_{NO}$

`=>`$n_O=\dfrac{3 \times 0.25}{2}=0.375(gam)$

Gộp thành $30+0.375 \times 16=36(gam)$

`=>`$n_{Fe_2O_3}=\dfrac{36}{160}=0.225(mol)$

Bảo toàn nguyên tố $Fe:$

$n_{Fe}=2n_{Fe_2O_3}=2 \times 0.225=0.45(mol)$

`=>`$m=0.45 \times 56=25.2(gam)$

`n_{NO} = 0,25` mol

Quy B thành `Fe(x ` mol`); O(y` mol`)`

`=> 56x + 16y = 30(1)`

Bảo toàn e

`=> 3n_{Fe} = 2n_O + 3n_{NO}`

`=> 3x = 2y + 0,25 . 3(2)`

`(1),(2) => x = 0,45; y = 0,3`

`=> m = m_{Fe} = 0,45 . 56 = 25,2g`