$Zn+2HCl\to ZnCl_2+H_2$
$Fe+2HCl\to FeCl_2+H_2$
$n_{HCl}=\frac{46.20\%}{36,5}=0,252 mol$
Theo PTHH, $n_{H_2}=0,5n_{HCl}=0,126 mol$
$m_{dd spứ}= m_{\text{kim loại}}+46-m_{H_2}$
$\Rightarrow m_{\text{kim loại}}= 169,4+0,126.2-46=123,652g$
Gọi a, b là mol Zn, Fe.
$\Rightarrow 65a+56b=123,652$ (1)
$n_{H_2}=0,126 mol$
$\Rightarrow a+b=0,126$ (2)
(1)(2) $\Rightarrow $ Nghiệm âm (??)