2Al + 6HCl \(\rightarrow\)2AlCl3 + 3H2 (1)
Fe + 2HCl \(\rightarrow\)FeCl2 + H2 (2)
nH2=\(\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Đặt nAl=a\(\Leftrightarrow m_{Al}=27a\)
nFe=b\(\Leftrightarrow m_{Fe}=56b\)
Ta có hệ pt:
\(\left\{{}\begin{matrix}27a+56b=5,5\\\dfrac{3}{2}a+b=0,2\end{matrix}\right.\)
Giải hệ ta có:
a=0,1;b=0,05
mAl=27.0,1=2,7(g)
C% Al=\(\dfrac{2,7}{5,5}.100\%=49\%\)
C% Fe=100-49=51%
b;\(\sum\)nHCl=0,1.3+0,05.2=0,4(mol)
CMHCl=\(\dfrac{0,4}{0,5}=0,8M\)
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