Đáp án:

`C\%_{Fe(NO_3)_3}\approx 13,13\%`

Giải thích các bước giải:

`n_{HNO_3}={242.31,5\%}/{63}=1,21(mol)`

`M_Z=10,125.4=40,5(g//mol)`

`->{n_{N_2O}}/{n_{NO}}={40,5-30}/{44-40,5}=3`

Mà `n_{N_2O}+n_{NO}={1,792}/{22,4}=0,08(mol)`

`->n_{N_2O}=0,06(mol);n_{NO}=0,02(mol)`

Quy X về `Fe:x(mol);Mg:y(mol);O:z(mol)`

`->56x+24y=16,96(1)`

Rắn sau cùng gồm `Fe_2O_3:0,5x(mol);MgO:y(mol)`

`->160.0,5x+40y=25,6(2)`

`(1)(2)->x=0,2;y=0,24`

Muối chứa:

`Fe(NO_3)_3:a(mol);Fe(NO_3)_2:b(mol);Mg(NO_3)_2:0,24(mol)`

Bảo toàn e:

`n_{NH_4NO_3}={1,21-0,06.10-0,02.4-2z}/{10}=0,053-0,2z(mol)`

`->m_Y=242a+180b+0,24.148+80(0,053-0,2z)=82,2(3)`

Bảo toàn Fe: `a+b=0,2(4)`

Bảo toàn e:

`3a+2b+2.0,24=0,02.3+0,06.8+8(0,053-0,2z)+2z(5)`

`(3)(4)(5)->a=0,14;b=0,06;z=0,14`

`->n_{Fe(NO_3)_3}=0,2(mol)`

`m_{dd\ spu}=16,96+0,14.16+242-0,08.40,5=257,96(g)`

`->C\%_{Fe(NO_3)_3}={0,14.242}/{257,96}.100\%\approx 13,13\%`