Hỗn hợp $X$ gồm $MgO$ ($x$ mol), $FeO$ ($y$ mol), $Fe_2O_3$ ($z$ mol). Quy đổi $Fe_3O_4=FeO.Fe_2O_3$.

$\Rightarrow 40x+72y+160z=5,92$   $(1)$

Bảo toàn $Mg$: $n_{Mg}=n_{MgO}=x (mol)$

Bảo toàn $Fe$: $n_{Fe}=n_{FeO}+2n_{Fe_2O_3}=y+2z (mol)$

$\Rightarrow 24x+56(y+2z)=4,16$    $(2)$

Sau các phản ứng, $6g$ rắn gồm:

$MgO$: $x$ mol

$Fe_2O_3$: $0,5y+z$ mol 

$\Rightarrow 40x+160(0,5y+z)=6$  $(3)$

$(1)(2)(3)\Rightarrow x=y=0,01; z=0,03$

Y gồm $MgCl_2$ ($0,01$ mol), $FeCl_2$ ($0,01$ mol), $FeCl_3$ ($0,06$ mol)

$n_{AgCl}=2n_{MgCl_2}+2n_{FeCl_2}+3n_{FeCl_3}=0,22(mol)$

$n_{Ag}=n_{FeCl_2}=0,01(mol)$

$\to m=0,0.108+0,22.143,5=32,65g$

PTHH và PT ion thu gọn:

$2Mg+O_2\buildrel{{t^o}}\over\to 2MgO$

$3Fe+2O_2\buildrel{{t^o}}\over\to Fe_3O_4$

$4Fe+3O_2\buildrel{{t^o}}\over\to 2Fe_2O_3$

$2Fe+O_2\buildrel{{t^o}}\over\to 2FeO$

$MgO+2HCl\to MgCl_2+H_2O$

$\Rightarrow MgO+2H^+\to Mg^{2+}+H_2O$

$Fe_3O_4+8HCl\to 2FeCl_3+FeCl_2+4H_2O$

$\Rightarrow Fe_3O_4+8H^+\to 2Fe^{3+}+Fe^{2+}+4H_2O$

$Fe_2O_3+6HCl\to 2FeCl_3+3H_2O$

$\Rightarrow Fe_2O_3+6H^+\to 2Fe^{3+}+3H_2O$

$FeO+2HCl\to FeCl_2+H_2O$

$\Rightarrow FeO+2H^+\to Fe^{2+}+H_2O$

$MgCl_2+2NaOH\to Mg(OH)_2+2NaCl$

$\Rightarrow Mg^{2+}+2OH^-\to Mg(OH)_2$

$FeCl_2+2NaOH\to Fe(OH)_2+2NaCl$

$\Rightarrow Fe^{2+}+2OH^-\to Fe(OH)_2$

$FeCl_3+3NaOH\to Fe(OH)_3+3NaCl$

$\Rightarrow Fe^{3+}+3OH^-\to Fe(OH)_3$

$Mg(OH)_2\buildrel{{t^o}}\over\to MgO+H_2O$

$2Fe(OH)_2+0,5O_2\buildrel{{t^o}}\over\to Fe_2O_3+2H_2O$

$2Fe(OH)_3\buildrel{{t^o}}\over\to Fe_2O_3+3H_2O$

$MgCl_2+2AgNO_3\to Mg(NO_3)_2+2AgCl$

$\Rightarrow Ag^++Cl^-\to AgCl$ 

$FeCl_2+3AgNO_3\to Fe(NO_3)_3+2AgCl+Ag$ 

$\Rightarrow Fe^{2+}+3Ag^++2Cl^-\to Fe^{3+}+2AgCl+Ag$ 

$FeCl_3+3AgNO_3\to Fe(NO_3)_3+3AgCl$

$\Rightarrow Ag^++Cl^-\to AgCl$ 

+ Ta có: $n_{O/X} = \frac{5,92 - 4,16}{16} = 0,11$ (mol)

$[O] + 2HCl → 2Cl + H_{2}O$ 

⇒$n_{Cl^{-}} = 2n_{O} = 0,22$ (mol)

+ Gọi $\left \{ {{n_{Mg} = x} \atop {n_{Fe} = y}} \right.$ ⇒$24x + 56y = 4,16$

+PTHH:  $Mg + \frac{1}{2}O_{2} → MgO$ 

                 $x$            $→$              $x$

                $2Fe + \frac{3}{2}O_{2} → Fe_{2}O_{3}$ 

                   $y$            $→$             $0,5y$

⇒$40x + 80y = 6$

+ Ta có: $\left \{ {{24x + 56y = 4,16} \atop {40x + 80y = 6}} \right.$ ⇔$\left \{ {{x = 0,01} \atop {y = 0,07}} \right.$ 

+ Áp dụng định luật bảo toàn e, ta có:

$2n_{Mg} + 3n_{Fe} = 2n_{O} + n_{Ag}$ 

⇔$n_{Ag} = 2n_{Mg} + 3n_{Fe} - 2n_{O} = 0,01$ (mol)

+ Kết tủa gồm $0,01$ mol $Ag$ và $0,22$ mol $AgCl$

⇒$m = 0,01.108 + 0,22.143,5 = 32,65$ (g).

BÀI THAM KHẢO. CHÚC EM HỌC TỐT. XIN HAY NHẤT.