$n_{H_2}=\dfrac{6,72}{22,4}=0,3mol \\a.PTHH : \\2Al+6HCl\to 2AlCl_3+3H_2↑ \\Fe+2HCl\to FeCl_2+H_2↑ \\Gọi\ n_{Al}=a;n_{Fe}=b \\Ta\ có : \\m_{hh}=27a+56b=11,1g \\n_{H_2}=1,5a+b=0,3mol \\Ta\ có\ hpt : \\\left\{\begin{matrix} 27a+56b=11,1 & \\ 1,5a+b=0,3 & \end{matrix}\right.⇔\left\{\begin{matrix} a=0,1 & \\ b=0,15 & \end{matrix}\right. \\⇒\%m_{Al}=\dfrac{0,1.27}{11,1}.100\%=24,32\% \\\%m_{Fe}=100\%-24,32\%=75,68\%$
b.Ảnh