Gọi số mol Mg, Al, Zn là a, b, c (mol)
=> 24a + 27b + 65c = 35 (1)
\(n_{H_2}=\dfrac{17,04}{22,4}=\dfrac{213}{280}\left(mol\right)\)
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
a------------------------>a
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
b---------------------->1,5b
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
c------------------------->c
=> 1,5b = 2a (2)
Và a + 1,5b + c = \(\dfrac{213}{280}\) (3)
(1)(2)(3) => \(\left\{{}\begin{matrix}a=\dfrac{809}{7560}\left(mol\right)\\b=\dfrac{809}{5670}\left(mol\right)\\c=\dfrac{277}{630}\left(mol\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{\dfrac{809}{7560}.24}{35}.100\%=7,338\%\\\%m_{Al}=\dfrac{\dfrac{809}{5670}.27}{35}.100\%=11,007\%\\\%m_{Zn}=\dfrac{\dfrac{277}{630}.65}{35}.100\%=81,655\%\end{matrix}\right.\)
