$n_{Fe}=16,8/56=0,3mol$
$PTHH :$
$Fe+2HCl\to FeCl_2+H_2$
$\text{Theo pt :}$
$n_{FeCl_2}=n_{H_2}=n_{Fe}=0,3mol$
$a/V_{H_2}=0,3.22,4=6,72l$
$b/m_{FeCl_2}=0,3.127=38,1g$
$\text{c/Theo pt :}$
$n_{HCl}=2.n_{Fe}=2.0,3=0,6mol$
$⇒C_{M_{HCl}}=\dfrac{0,6}{0,3}=2M$