$n_{HCl}=0,1.2=0,2mol$
$a.Fe+2HCl\to FeCl_2+H_2↑$
$\text{b.Theo pt :}$
$n_{Fe}=1/2.n_{HCl}=1/2.0,2=0,1mol$
$⇒m=m_{Fe}=0,1.56=5,6g$
$\text{c.Theo pt :}$
$n_{H_2}=1/2.n_{HCl}=1/2.0,2=0,1mol$
$⇒V=V_{H_2}=0,1.22,4=2,24l$
$d.m_{dd\ HCl}=1,05.100=105g$
$\text{Theo pt :}$
$n_{FeCl_2}=1/2.n_{HCl}=1/2.0,2=0,1mol$
$⇒m_{FeCl_2}=0,1.127=12,7g$
$m_{dd\ spu}=5,6+105-0,1.2=110,4g$
$⇒C\%_{FeCl_2}=\dfrac{12,7}{110,4}.100\%=11,5\%$