a,
$Zn+2HCl\to ZnCl_2+H_2$
$Fe+2HCl\to FeCl_2+H_2$
$n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)$
$\Rightarrow n_{HCl}=2n_{H_2}=0,6(mol)$
$\to x=\dfrac{0,6.36,5.100}{200}=10,95\%$
b,
Gọi x, y là số mol Zn, Fe.
$\Rightarrow 65x+56y=18,6$ (1)
$n_{H_2}=x+y=0,3$ (2)
(1)(2)$\Rightarrow x=0,2; y=0,1$
$\%m_{Zn}=\dfrac{0,2.65.100}{18,6}=69,89\%$
$\%m_{Fe}=30,11\%$