\(\begin{array}{l}
Dat\,(n_{Mg};n_{Fe};n_{Zn})=(x;y;z)(mol)\\
\to 24x+56y+65z=46,1(1)\\
n_{H_2}=\frac{17,92}{22,4}=0,8(mol)\\
Mg+2HCl\to MgCl_2+H_2(*)\\
Fe+2HCl\to FeCl_2+H_2(**)\\
Zn+2HCl\to ZnCl_2+H_2\\
Theo\,PT:\,x+y+z=n_{H_2}=0,8(2)\\
Ta\,co\,2n_{H_2(*)}=n_{H_2(**)}\\
\to 2x=y(3)\\
(1)(2)(3)\to x=0,1;y=0,2;z=0,5\\
\to \%m_{Mg}=\frac{0,1.24}{46,1}.100\%≈5,2\%\\
\to \%m_{Fe}=\frac{0,2.56}{46,1}.100\%≈24,3\%\\
\to \%m_{Zn}=100-5,2-24,3=70,5\%
\end{array}\)

$46,1(g)\begin{cases}Mg:x\\Fe:2x\\Zn:y\end{cases}\xrightarrow{\quad HCl\quad }17,92(l)H_2$

$m_{hh}=24x+56.2x+65y=46,1$

$n_{H_2}=x+2x+y=\dfrac{17,92}{22,4}=0,8$

$\to \begin{cases}n_{Mg}=0,1\ (mol)\\n_{Fe}=0,2\ (mol)\\n_{Zn}=0,5\ (mol)\end{cases}$

$\%m_{Mg}=\dfrac{0,1.24}{46,1}.100=5,2\%$

$\%m_{Fe}=\dfrac{0,2.56}{46,1}.100=24,3\%$

$\%m_{Zn}=100\%-5,2\%-24,3\%=70,5\%$