Đặt \(n_{Mg}=a(mol);n_{Zn}=b(mol)\Rightarrow 24a+65b=15,4(1)\)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)\\ PTHH:Mg+2HCl\to MgCl_2+H_2\\ Zn+2HCl\to ZnCl_2+H_2\\ \Rightarrow x+y=0,3(2)\\ (1)(2)\Rightarrow x=0,1(mol);y=0,2(mol)\\ \Rightarrow \%_{Mg}=\dfrac{0,1.24}{15,4}.100\%=15,58\%\)