$a,\text{Đặt KL trung bình là R}\rightarrow R(II)\\ PTHH:R+2HCl\to RCl_2+H_2\\ \Rightarrow n_{H_2}=n_R=\dfrac{6,72}{22,4}=0,3(mol)\\ \Rightarrow M_R=\dfrac{8,8}{0,3}=29,33(g)$

Vậy 2 KL đó là Mg (24) và Ca (40)

Đặt $(n_{Mg};n_{Ca})=(x;y)(mol)$

$PTHH:Mg+2HCl\to MgCl_2+H_2(1)\\ Ca+2HCl\to CaCl_2+H_2(2)\\ \Rightarrow \begin{cases} x+y=n_{H_2}=0,3\\ 24y+40y=m_{hh}=8,8 \end{cases}\Rightarrow \begin{cases} x=0,2(mol)\\ y=0,1(mol) \end{cases}\\ \Rightarrow \begin{cases} \%m_{Mg}=\dfrac{0,2.24}{8,8}.100\%=54,545\%\\ \%m_{Ca}=100\%-54,545\%=45,455\% \end{cases}\\ b,n_{H_2(2)}=n_{CaCl_2}=y=0,1(mol);n_{H_2(1)}=n_{MgCl_2}=x=0,2(mol)\\ \Rightarrow m_{CaCl_2}=0,1.111=11,1(g);m_{MgCl_2}=0,2.95=19(g)\\ m_{Mg}=0,2.24=4,8(g);m_{Ca}=0,1.40=4(g)\\ \Rightarrow \begin{cases} C\%_{MgCl_2}=\dfrac{19}{4,8+200-0,2.2}.100\%=9,295\%\\ C\%_{CaCl_2}=\dfrac{11,1}{4+200-0,1.2}.100\%=5,45\% \end{cases}$