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

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

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

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

Bảo toàn khối lượng:
$m_Y+m_{O_2}+m_{Cl_2}=m_{Z}$

`=>`$m_{O_2}+m_{Cl_2}=25.36$ 

`<=>`$32n_{O_2}+71n_{Cl_2}=25.36$ $(1)$

$n_X=0.5(mol)$

Lại có: $n_{O_2}+n_{Cl_2}=0.5$ $(2)$

Từ $(1)$ và $(2)$ suy ra: $n_{O_2}=0.26(mol);n_{Cl_2}=0.24(mol)$

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

 $x$   `->`   $0.5x$   `->`   $x$

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

 $y$   `->`   $0.75y$  `->` $0.5y$

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

 $z$   `->`   $z$   `->`   $z$

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

 $t$   `->`   $1.5t$   `->`  $t$

`->`$\begin{cases} 24(x+z)+27(y+t)=16.98\\0.5x+0.75y+z+1.5t=0.5\\40x+51y+95z+130.5t=42.34\\z+1.5t=0.24 \end{cases}$

`=>`$\begin{cases} x=0.31\\b=0.14\\c=0.24\\t=0 \end{cases}$

$\%m_{Al}=\dfrac{0.14 \times 27}{16.98} \times 100\%=22.26\%$

$\%m_{Mg}=77.74\%$

Đáp án:

$\rm n_{X}=\dfrac{12,395}{24,79}=0,5\ (mol)$

Bảo toàn khối lượng:

$\rm m_{X}=42,34-16,98=25,36\ (g)$

Hệ phương trình:

$\begin{cases}\rm 71n_{Cl_2}+32n_{O_2}=25,36\\\rm n_{Cl_2}+n_{O_2}=0,5\end{cases}\to\begin{cases}\rm n_{Cl_2}=0,24\\\rm b=0,26\end{cases}$

Phương trình phản ứng:

$\rm 2Mg+O_2 \xrightarrow{t^°}2MgO\\\kern,3cma\kern,0cm\to0,5a\kern,2cm\to\kern,2cma\kern1cm(mol)\\4Al+{3}O_2 \xrightarrow{t^°}2Al_2O_3\\\kern,2cmb\kern,0cm\to0,75b\kern,1cm\to\kern,1cm0,5\kern,55cm(mol)\\Mg+Cl_2 \xrightarrow{t^°}MgCl_2\\\kern,2cmc\kern,15cm\to\kern,15cm c\kern,4cm\to\kern,4cmc\kern1,2cm(mol)\\2Al+{3}Cl_2 \xrightarrow{t^o}2AlCl_3\\\kern,2cmd\kern,0cm\to1,5d\kern,25cm\to\kern,3cmd\kern,9cm(mol)$

Hệ phương trình:

$\begin{cases}\rm 24a+27b+24c+27d=16,98\\\rm 40a+51b+95c+130,5d=42,34\\\rm 0,5a+0,75b+c+1,5d=0,5\\\rm 0,5a+0,75b=0,26\end{cases}\to\begin{cases}\rm a=0,31\\\rm b=0,14\\\rm c=0,24\\\rm d=0\end{cases}$

$\rm\%m_{Mg}=\dfrac{(0,31+0,24)×24}{16,98}×100\%=77,74\%\\\%m_{Al}=(100-77,74)\%=22,26\%$