2Al + 6HCl \(\rightarrow\)2AlCl3 + 3H2 (1)
Fe + 2HCl \(\rightarrow\)FeCl2 + H2 (2)
nHCl=\(\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Đặt nAl=a
nFe=b
Ta có hệ:
\(\left\{{}\begin{matrix}27a+56b=11\\\dfrac{3}{2}a+b=0,4\end{matrix}\right.\)
=>a=0,2;b=0,1
mAl=27.0,2=5,4(g)
%mAl=\(\dfrac{5,4}{11}.100\%=49,1\%\)
%mFe=100-49,1=50,9%
b;
Theo PTHH 1 và 2 ta có:
3nAl=nHCl=0,6(mol
2nFe=nHCl=0,2(mol)
Vdd HCl=\(\dfrac{0,6+0,2}{2}=0,4\left(lít\right)\)