$\begin{cases} Fe_aO_b\\MgO\\ \end{cases} \xrightarrow{+ HCl} \begin{cases} MgCl_2\\FeCl_{{2b}/a} \end{cases}$

$\xrightarrow{ + NaOH}$$\begin{cases} Fe(OH)_{{2b}/a}\\Mg(OH)_2 \end{cases}$$\xrightarrow{t°}$$\begin{cases} Fe_2O_3\\MgO \end{cases}$$\xrightarrow{+H_2}$$\begin{cases} Fe\\MgO\end{cases}$

`n_O (Fe_2O_3) = {4,4 - 3,68}/{16} = 0,045(mol)`

Bảo toàn `O`: 

`n_{Fe_2O_3} = 1/3n_O = {0,045}/3 = 0,015(mol)`

Bảo toàn `Fe`: 

`n_{Fe} = 2n_{Fe_2O_3} = 0,015.  2 = 0,03(mol)`

`n_{MgO} (r) = {3,68 - 0,03.  56}/{40} = 0,05(mol)`

Bảo toàn `Mg`: 

`n_{MgO} (X) = n_{MgO}(r) = 0,05(mol)`

`n_O (Fe_xO_y) = {4,32 - 0,03 . 56 - 0,05 . 40}/{16} = 0,04(mol)`

`a/b = {n_{Fe}}/{n_O} = {0,03}/{0,04} = 3/4`

`→`  Oxit sắt là `Fe_3O_4` 

`%m_{MgO} = {0,05 . 40}/{4,32} .100= 46,3%`

`%m_{Fe_3O_4} = 100 - 46,3 = 53,7%`

Đáp án:

$CTHH:\ Fe_3O_4\\ \%m_{MgO}=46,3\%\\ \%m_{Fe_3O_4}=53,7\%$

Giải thích các bước giải:

$Oxit\ sắt:\ Fe_2O_n\\ m_{O\ (Fe_2O_3)}=m_M-m_{\text{chất rắn}}\\=4,4-3,68\\=0,72(g)\\ n_{O\ (Fe_2O_3)}=\dfrac{0,72}{16}=0,045(mol)\\ \xrightarrow{\text{BTNT O}}\ n_{Fe_2O_3}=\dfrac{1}{3}.n_{O\ (Fe_2O_3)}=0,015(mol)\\ m_{Fe_2O_3}==0,015.160=2,4(g)\\ m_{MgO\ (M)}=4,4-2,4=2(g)\\ n_{MgO\ (M)}=\dfrac{2}{40}=0,05(mol)\\ \xrightarrow{\text{BTNT Mg}}\ n_{MgO\ (X)}=n_{MgO\ (M)}=0,05(mol)\\ m_{MgO\ (X)}=0,05.40=2(g)\\ m_{Fe_2O_n}=4,32-2=2,32(g)\\ \xrightarrow{\text{BTNT Fe}}\ n_{Fe_2O_n}=n_{Fe_2O_3}=0,015(mol)\\ \Rightarrow \dfrac{2,32}{112+16n}=0,015\\ \Rightarrow n=\dfrac{8}{3}\\ \to CTHH:\ Fe_3O_4$

$\%m_{MgO}=\dfrac{2}{4,32}.100\%=46,3\%\\ \%m_{Fe_3O_4}=\dfrac{2,32}{4,32}.100\%=53,7\%$