$2Fe+3Cl2\to 2FeCl3$
$2Al+3Cl2\to 2AlCl3$
Gọi $n_{Fe}=x ; n_{Al}=y$
Ta có
$56x+27y=8,3$
$162,5x+133,5y=29,6$
Giải hệ ta đc:
$x=y=0,1$
$\%mFe=\dfrac{56.0,1}{8,3}.100\%=67,47\%$
$\%mAl=100\%-67,47\%=32,53\%$
$n_{Cl_2}=\dfrac{0,1.3}{2}+\dfrac{0,1.3}{2}=0,3mol$
$⇒V_{Cl_2}=0,3.22,4=6,72 l$