a,

$2Fe+O_2\to 2FeO$

$4Fe+3O_2\to 2Fe_2O_3$

$3Fe+2O_2\to Fe_3O_4$

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

$3FeO+10HNO_3\to 3Fe(NO_3)_3+NO+5H_2O$

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

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

b,

Gọi chung hh $A$ là $Fe_xO_y$

$n_{Fe}=\dfrac{11,2}{56}=0,2(mol)$

$2xFe+yO_2\to 2Fe_xO_y$

$\to n_{Fe_xO_y}=\dfrac{0,2}{x}(mol)$

$3Fe_xO_y+(12x-2y) HNO_3\to 3xFe(NO_3)_3+(3x-2y)NO+(6x-y)H_2O$

$n_{NO}=\dfrac{0,896}{22,4}=0,04(mol)$

$\to \dfrac{0,04.3}{3x-2y}=\dfrac{0,2}{x}$

$\to 0,12x=0,2(3x-2y)$

$\to 0,4y=0,48x$

$\to \dfrac{x}{y}=\dfrac{5}{6}$

Do đó trong $A$, $\%m_{Fe}=\dfrac{5.56.100}{5.56+6.16}=\dfrac{3500}{47}\%$

$\to m_1=11,2:\dfrac{3500}{47}\%=15,04g$

$n_{Fe(NO_3)_3}=xn_{Fe_xO_y}=0,2(mol)$

$\to m_2=0,2.242=48,4g$

`n_{Fe}=\frac{11,2}{56}=0,2(mol)`

`a,`

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

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

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

`3FeO+10HNO_3->3Fe(NO_3)_3+NO+5H_2O`

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

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

`b,`

`n_{NO}=\frac{0,896}{22,4}=0,04(mol)`

Coi hỗn hợp oxit gồm $\begin{cases}Fe : \ 0,2(mol)\\ O : \ x(mol)\\\end{cases}$

Sơ đồ cho - nhận e

\(\rm \mathop{Fe}\limits^{+0}\to \mathop{Fe}\limits^{+3}+3e \\\mathop{O}\limits^{+0}+2e\to \mathop{O}\limits^{-2} \\\mathop{N}\limits^{+5}+3e\to \mathop{N}\limits^{+2}\)

BTe

`->3n_{Fe}=2n_{O}+3n_{NO}`

`->2x+3.0,04=0,6`

`->x=0,24(mol)`

`->m_{1}=0,2.56+0,24.16=15,04(g)`

`b,`

Bảo toàn `Fe`

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

`->m_2=0,2.242=48,4(g)`