$a,PTPƯ:Fe+2HCl\xrightarrow{} FeCl_2+H_2↑$
$n_{Fe}=\dfrac{22,4}{56}=0,4mol.$
$Theo$ $pt:$ $n_{H_2}=n_{Fe}=0,4mol.$
$⇒V_{H_2}=0,4.22,4=8,96l.$
$b,Theo$ $pt:$ $n_{HCl}=2n_{Fe}=0,8mol.$
$⇒V_{HCl}=\dfrac{0,8}{0,5}=1,6l.$
$c,Theo$ $pt:$ $n_{FeCl_2}=n_{Fe}=0,4mol.$
$⇒CM_{FeCl_2}=\dfrac{0,4}{1,6}=0,25M.$
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