Đáp án `+` Giải thích các bước giải:

PTHH:
$2Mg + O_2 \xrightarrow{t^o} 2MgO(1)$

$4Al + 3O_2 \xrightarrow{t^o}  2Al_2O_3(2)$

$Mg + Cl_2 \xrightarrow{t^o}  MgCl_2(3)$

$2Al + 3Cl_2 \xrightarrow{t^o}  2AlCl_3(4)$

$MgO + 2HCl \to MgCl_2 + H_2O(5)$

$Al_2O_3 + 6HCl \to 2AlCl_3 + 3H_2O(6)$

$MgCl_2 + 2AgNO_3 \to Mg(NO_3)_2 + 2AgCl \downarrow(7)$

$AlCl_3 + 3AgNO_3 \to Al(NO_3)_3 + 3AgCl \downarrow(8)$

$n_{HCl} = 0,2.0,2 = 0,04 (mol); n_{AgCl} = \dfrac{11,48}{143,5} = 0,08 (mol)$

Theo PT $(5), (6): n_O = \dfrac{1}{2} n_{HCl} = 0,02 (mol)$

$\text{Bảo toàn nguyên tố O}: n_{O_2} = \dfrac{1}{2} n_O = 0,01 (mol)$

$\text{Bảo toàn nguyên tố Cl}: 2n_{Cl_2} + n_{HCl} = n_{AgCl}$

$\rightarrow n_{Cl_2} = \dfrac{0,08 - 0,04}{2} = 0,02 (mol)$

Vì tỉ lệ thể tích cũng là tỉ lệ số mol nên ta có:

$\%V_{O_2} = \%n_{O_2} = \dfrac{0,01}{0,01+0,02} .100\% = 33,33\%$

$\%V_{Cl_2} = 100\% - 33,33\% = 66,67\%$

Gọi $n_{Mg} = x (mol); n_{Al} = y (mol)$

$\rightarrow 24x + 27y = 0,78 (*)$

Quá trình oxi hoá - khử:

$Al^0 \rightarrow Al^{+3} + 3e$

$Mg^0 \rightarrow Mg^{+2} + 2e$

$2Cl^0 +2e \rightarrow 2Cl^{-1}$

$2O^0 + 4e \rightarrow 2O^{-2}$

$BTelectron: 3n_{Al} + 2n_{Mg} = 2n_{Cl_2} + 4n_{O_2}$

$\rightarrow 3y + 2x = 2.0,02 + 4.0,01 = 0,08 (mol) (**)$

Từ $(*), (**) \rightarrow x = 0,01; y = 0,02$

$\rightarrow \%m_{Al} = \dfrac{0,02.27}{0,78} .100\% = 69,23\%$

$\rightarrow \%m_{Mg} = 100\% - 69,23\% = 30,77\%$

$\rm 0,78(g)\ hh\ E : $ $\begin{cases} \rm Mg : a(mol)\\\rm Al : b(mol) \end{cases}$

`=>`$\rm m_E=24a+27b=0,78\ (1)$

$\rm hh\ B : $ $\begin{cases} \rm MgO\\\rm Al_2O_3\\\rm MgCl_2\\\rm AlCl_3 \end{cases}$

$\rm 200(ml)=0,2(l)$

$\rm n_{HCl}=0,2.0,2=0,04(mol)$

$\rm MgO+2HCl\to MgCl_2+H_2O$

$\rm Al_2O_3+6HCl\to 2AlCl_3+3H_2O$

$\xrightarrow{\text{BTNT :}}$ $\rm n_O=n_{H_2O}=\dfrac{1}{2}n_{HCl}=0,02(mol)$

$\xrightarrow{\text{BTNT O :}}$ $\rm n_{O_2}=\dfrac{n_{O}}{2}=0,01(mol)$

$\rm dd\ D : $ $\begin{cases} \rm MgCl_2\\\rm AlCl_3 \end{cases}$

$\rm n_{AgCl}=\dfrac{11,48}{143,5}=0,08(mol)$

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

$\rm AlCl_3+3AgNO_3\to 3AgCl+Al(NO_3)_3$

$\xrightarrow{\text{BTNT Cl :}}$ $\rm 2n_{Cl_2}+n_{HCl}=n_{AgCl}$

`<=>`$\rm 2n_{Cl_2}+0,04=0,08$

`=>`$\rm n_{Cl_2}=0,02(mol)$

Ở cùng điều kiện: `%`$\rm n$`=%`$\rm V$

`%`$\rm V_{Cl_2}$`={0,02}/{0,02+0,01}.100%~~66,67%`

`%`$\rm V_{O_2}$`=100%-66,67%=33,33%`

Quá trình nhường - nhận electron:

\begin{array}{c|c|c} \rm Mg^o\to Mg^{+2}+2e&\rm 2O^{o}+4e\to 2O^{-2}\\ \rm Al^o\to Al^{+3}+3e& \rm 2Cl^o+2e\to 2Cl^{-1}\end{array}

$\xrightarrow{\text{BTe :}}$ $\rm 2n_{Mg}+3n_{Al}=4n_{O_2}+2n_{Cl_2}$

`<=>`$\rm 2a+3b=4.0,01+2.0,02$

`=>`$\rm 2a+3b=0,08\ (2)$

`(1)(2)=>`$\rm a=0,01(mol)\ , b=0,02(mol)$

`%`$\rm m_{Mg}$`={0,01.24}/{0,78}.100%~~30,77%`

`%`$\rm m_{Al}$`=100%-30,77%=69,23%`