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

$a$   `->` $a$   `->` $a$

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

$b$   `->` $\dfrac{2b}{3}$   `->` $\dfrac{2b}{3}$

$\dfrac{1}{2}O_2+Mg \xrightarrow{t^o} MgO$

$c$   `->` $2c$   `->` $2c$ 

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

$d$  `->`  $\dfrac{4d}{3}$ `->` $\dfrac{2d}{3}$

$n_X=\dfrac{3.92}{22.4}=0.175(mol)$

`->`$n_{Cl_2}+n_{O_2}=0.175$

`->`$n_{Cl_2}=0.175-n_{O_2}$

Bảo toàn khối lượng:

$m_{Cl_2}+m_{O_2}+m_{Mg+Al}=m_{Z}$

`=>`$71 \times (0.175-n_{O_2})+32n_{O_2}=15.05-5.55=9.5$ 

`=>`$n_{O_2}=0.075(mol)$

`=>`$n_{Cl_2}=0.175-0.075=0.1(mol)$

$n_{Al}=\dfrac{2b}{3}+\dfrac{4d}{3}(mol)$

$n_{Mg}=(0.1-b)+2 \times (0.075-d)(mol)$

`->`$27(\dfrac{2b}{3}+\dfrac{4d}{3})+24[(0.1-b)+2(0.075-d)]=5.55$

`=>`$b=\dfrac{-12d+0.45}{6}$

Vì không biết tỷ lệ phản ứng ta chọn giá trị bất kỳ cho $d$ sao cho $b>0$

Do đó, ta chọn $d=0.035$`=>`$b=0.005(mol)$

`=>`$a=0.095;c=0.04$

`=>`$n_{Mg}=a+2c=0.175(mol)$

`=>`$\%m_{Mg}=\dfrac{0.175 \times 24}{5.55} \times 100\%=75.68\%$

`Mg+Cl_2` $\xrightarrow{t^o}$ `MgCl_2`

`2Al+3Cl_2` $\xrightarrow{t^o}$ `2AlCl_3`

`2Mg+O_2` $\xrightarrow{t^o}$ `2MgO`

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

BTKL: `71n_{Cl_2}+32n_{O_2}=15,05-5,55=9,5(1)`

`n_{Cl_2}+n_{O_2}=n_X={3,92}/{22,4}=0,175(2)`

`(1)(2)->n_{Cl_2}=0,1(mol);n_{O_2}=0,075(mol)`

`Y:Mg:x(mol);Al:y(mol)`

`24x+27y=5,5(3)`

Bte: `2x+3y=2n_{Cl_2}+4n_{O_2}=0,5(2)`

`(1)(2)->x=1/6;y=1/{18}`

`\%m_{Mg}={1/{6}.24}/{5,5}.100\%=72,73\%`