$a,PTPƯ:Fe+2HCl\xrightarrow{} FeCl_2+H_2↑$
$n_{Fe}=\dfrac{5,6}{56}=0,1mol.$
Đổi 100 ml = 0,1 lít.
$n_{HCl}=0,1.1=0,1mol.$
$\text{Lập tỉ lệ:}$ $\dfrac{0,1}{1}>\dfrac{0,1}{2}$
$⇒Fe$ $dư.$
$Theo$ $pt:$ $n_{H_2}=\dfrac{1}{2}n_{HCl}=0,05mol.$
$⇒V_{H_2}=0,05.22,4=1,12l.$
$b,Fe$ $dư.$
$n_{Fe}(dư)=0,1-\dfrac{0,1}{2}=0,05mol.$
$⇒m_{Fe}(dư)=0,05.56=2,8g.$
$c,Theo$ $pt:$ $n_{FeCl_2}=\dfrac{1}{2}n_{HCl}=0,05mol.$
$⇒CM_{FeCl_2}=\dfrac{0,05}{0,1}=0,5M.$
chúc bạn học tốt!