Cách 1:

`n_{Fe\ pu}=x(mol);n_{Fe_3O_4}=y(mol)`

`->56x+232y=18,5-1,46=17,04(1)`

`Fe+4HNO_3->Fe(NO_3)_3+NO+2H_2O(*)`

`3Fe_3O_4+28HNO_3->9Fe(NO_3)_3+NO+14H_2O`

`y->{28}/{3}y->3y->y/3`

`Fe+2Fe(NO_3)_3->3Fe(NO_3)_2(**)`

`->n_{Fe\ (*)}=0,1-y/3(mol)`

`->n_{Fe\ (**)}=x-(0,1-y/3)=x+y/3-0,1(mol)`

`n_{Fe(NO_3)_3}=0,1-y/3+3y=0,1+{8}/{3}y(mol)`

`->(x+y/3-0,1).2=0,1+{8}/{3}y`

`->2x+{2y}/{3}-0,2=0,1+{8}/{3}y`

`->2x-2y=0,3(2)`

`(1)(2)

Cách 2:

`n_{Fe\ pu}=x(mol);n_{Fe_3O_4}=y(mol)`

`->56x+232y=18,5-1,46=17,04(1)`

`Fe->Fe^{+2}+2e`

`Fe_3^{+8//3}+2e->3Fe^{+2}`

`N^{+5}+3e->N^{+2}`

Bte: `2x=2y+3n_{NO}=2y+0,3(2)`

`(1)(2)->x=0,18;y=0,03`

`->n_O=4y=0,12(mol)`

Bte: `n_{HNO_3}=2n_O+4n_{NO}=0,64(mol)`