Gọi số mol Al, Zn, Fe là a, b, c (mol)
=> 27a + 65b + 56c = 14,7
Và \(27a=\dfrac{18}{31}\left(65b+56c\right)\)
=> \(\left\{{}\begin{matrix}a=0,2\left(mol\right)\\65b+56c=9,3\left(g\right)\end{matrix}\right.\)
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)
Theo PTHH: \(n_{H_2}=0,2.1,5+b+c=0,45\)
=> b + c = 0,15
=> b = 0,1 (mol); c = 0,05 (mol)
\(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{27.0,2}{14,7}.100\%=36,73\%\\\%m_{Zn}=\dfrac{0,1.65}{14,7}.100\%=44,22\%\\\%m_{Fe}=\dfrac{0,05.56}{14,7}.100\%=19,05\%\end{matrix}\right.\)
