$n_{Fe}=22,4/56=0,4mol$
$PTHH : Fe+2HCl\to FeCl_2+H_2↑$
a,Theo pt :
$n_{H_2}=n_{fe}=0,4mol$
$⇒V_{H_2}=0,4.22,4=8,96l$
b.Theo pt :
$n_{HCl}=2.n_{Fe}=2.0,4=0,8mol$
$⇒m_{HCl}=0,8.36,5=29,2g$
$⇒m_{ddHCl}=\dfrac{29,2}{15\%}≈194,66g$
$c,m_{ddspu}=22,4+194,66-0,4.2=216,26g$
$⇒C\%_{FeCl_2}=\dfrac{0,2.127}{216,26}.100\%=58,73\%$