2Al + 6HCl -> 2AlCl3 + 3H2 (1)
Fe +2HCl ->FeCl2 + H2 (2)
nH2=\(\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Đặt nAl=a
nFe=b
Ta có heek pt:
\(\left\{{}\begin{matrix}27a+56b=11\\1,5a+b=0,4\end{matrix}\right.\)
=>a=0,2;b=0,1
mFe=56.0,1=5,6(g)
%mFe=\(\dfrac{5,6}{11}.100\%=51\%\)
%mAl=100-51=49%