$n_{HCl}=\dfrac{400.5,475\%}{36,5}=0,6(mol)$
$n_{H_2}=\dfrac{5,6}{22,4}=0,25(mol)$
$Fe+2HCl\to FeCl_2+H_2$
$\Rightarrow n_{FeCl_2}=n_{Fe}=0,25(mol)$
$m_{Fe}=0,25.56=14g$
$n_{HCl\text{dư}}=0,6-0,25.2=0,1(mol)$
Ta có $m_{dd\text{spu}}=14+400-0,25.2=413,5g$
$C\%_{HCl}=\dfrac{0,1.36,5.100}{413,5}=0,88\%$
$C\%_{FeCl_2}=\dfrac{0,25.127.100}{413,5}=7,68\%$