$n_{Fe}=2,8/56=0,05mol$
$a/Fe+2HCl\to FeCl_2+H_2↑$
$\text{b/Theo pt :}$
$n_{HCl}=2.n_{Fe}=2.0,05=0,1mol$
$⇒m_{HCl}=0,1.36,5=3,65g$
$⇒C\%_{HCl}=\dfrac{3,65}{50}.100\%=7,3\%$
$\text{c/Theo pt :}$
$n_{FeCl_2}=n_{Fe}=0,05mol$
$⇒m_{FeCl_2}=0,05.127=6,35g$
$m_{ddspu}=2,8+50-0,05.2=52,7g$
$⇒C\%_{FeCl_2}=\dfrac{6,35}{52,7}.100\%=12,05\%$