a)
$Fe + 2HCl \to FeCl_2 + H_2$
$Zn + 2HCl \to ZnCl_2 + H_2$
b)
Gọi $n_{Fe} = a(mol) ; n_{Zn} = b(mol) \Rightarrow 56a + 65b = 12,1(1)$
Theo PTHH : $n_{H_2} = a + b = \dfrac{4,48}{22,4} = 0,2(2)$
Từ (1)(2) suy ra : a = b = 0,1
$\%m_{Fe} = \dfrac{0,1.56}{12,1}.100\% = 46,3\%$
$\%m_{Zn} = 100\% - 46,3\% = 53,7\%$