a.2Al + 6HCl -> 2AlCl3 + 3H2
2a 6a 3a
Fe + 2HCl -> FeCl2 + H2
b 2b b
b.\(nH2=\dfrac{1.344}{22.4}=0.06mol\)
Ta có hệ phương trình: \(\left\{{}\begin{matrix}54a+56b=2.22\\3a+b=0.06\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0.01mol\\b=0.03mol\end{matrix}\right.\)
\(\%mAl=\dfrac{0.01\times2\times27\times100}{2.22}=24.3\%\)
%mFe = 100 - 24.3 =75.7%
c.nHCl = 6a + 2b = 0.12 mol
\(CM_{HCl}=\dfrac{0.12}{0.2}=0.6M\)
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