$n_{Fe}=2,8/56=0,05mol$
$a/Fe+2HCl\to FeCl_2+H_2$
$b/m_{dd HCl}=50.1,18=59g$
Theo pt :
$n_{HCl}=2.n_{Fe}=2.0,05=0,1mol$
$⇒m_{HCl}=0,1.36,5=3,65g$
$⇒C\%_{HCl}=\dfrac{3,65}{59}.100\%=6,19\%$
c/Theo pt :
$n_{FeCl_2}=n_{H_2}=n_{Fe}=0,05mol$
$⇒m_{FeCl_2}=0,05.127=6,35g$
$m_{dd spư}=2,8+59-0,05.2=61,7g$
$⇒C\%_{FeCl_2}=\dfrac{6,35}{61,7}.100\%=10,29\%$