`Fe+2HCl->FeCl_2+H_2`
`2Al+6HCl->2AlCl_3+3H_2`
`n_{Fe;Al}=x;y(mol)`
`n_{H_2}=(17,92)/(22,4)=0,8(mol)`
`n_{H_2}=x+3/2y=0,8`
`m_{hh}=56x+27y=22`
$\begin{cases} x+\dfrac{3}{2}y=0,8\\56x+27y=22\end{cases}$
$\begin{cases} x=0,8-\dfrac{3}{2}y\\56.(0,8-\dfrac{3}{2}y)+27y=22\end{cases}$
$\begin{cases} x=0,8-\dfrac{3}{2}y\\44,8-84y+27y=22\end{cases}$
$\begin{cases} x=0,8-\dfrac{3}{2}y\\44,8-57y=22\end{cases}$
$\begin{cases} x=0,8-\dfrac{3}{2}y\\y=0,4\end{cases}$
$\begin{cases} x=0,8-\dfrac{3}{2}.0,4\\y=0,4\end{cases}$
$\begin{cases} x=0,2\\y=0,4\end{cases}$
`%m_{Fe}=(0,2.56)/(22%)=50,91%`
`%m_{Al}=100%-50,91%=49,09%`