a) Fe+2HCl---->FeCl2+H2
x----------2x----------------x
2Al+6HCl---->2AlCl3+3H2
y--------3y-----------------1,5y
b)Ta có
n H2=\(\frac{3,024}{22,4}=0,135\left(mol\right)\)
Theo bài ra ta có hệ pt
\(\left\{{}\begin{matrix}56x+27y=4,14\\x+1,5y=0,135\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,045\\y=0,06\end{matrix}\right.\)
%m Fe=\(\frac{0,045.56}{4,14}.100\%=60,87\%\)
%m Al =100-60,87=39,13%
c)Theo pthh
n\(_{HCl}=2n_{H2}=0,27\left(mol\right)\)
m dd HCl =\(\frac{0,27.36,5.100}{10}=91,25\left(g\right)\)
