a,
$Fe+2HCl\to FeCl_2+H_2$
$Cu+Cl_2\buildrel{{t^o}}\over\to CuCl_2$
$2Fe+3Cl_2\buildrel{{t^o}}\over\to 2FeCl_3$
b,
$n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)$
$\Rightarrow n_{Fe}=n_{H_2}=0,1(mol)$
$n_{Cl_2}=\dfrac{6,72}{22,4}=0,3(mol)$
Ta có $1,5n_{Fe}+n_{Cu}=0,3$
$\Rightarrow n_{Cu}=0,15(mol)$
$\%m_{Fe}=\dfrac{0,1.56.100}{0,1.56+0,15.64}=36,84\%$
$\%m_{Cu}=63,16\%$
