$a\big)Fe+2HCl\to FeCl_2+H_2$
$b\big)$
$n_{Fe}=\dfrac{11,2}{22,4}=0,2(mol)$
$\to n_{HCl}=2n_{Fe}=0,4(mol)$
$\to m_{dd\,HCl}=\dfrac{0,4.36,5}{14,6\%}=100(g)$
$c\big)$
$n_{FeCl_2}=n_{H_2}=n_{Fe}=0,2(mol)$
$m_{dd\,spu}=11,2+100-0,4.2=110,8(g)$
$\to C\%_{FeCl_2}=\dfrac{0,2.127}{110,8}.100\%\approx 22,92\%$