$a,PTPƯ:Fe+2HCl\xrightarrow{} FeCl_2+H_2↑$
$n_{Fe}=\dfrac{22,4}{56}=0,4mol.$
$Theo$ $pt:$ $n_{H_2}=n_{Fe}=0,4mol.$
$⇒V_{H_2}=0,4.22,4=8,96l.$
$b,Theo$ $pt:$ $n_{FeCl_2}=n_{Fe}=0,4mol.$
$⇒m_{FeCl_2}=0,4.127=50,8g.$
$c,Theo$ $pt:$ $n_{HCl}=2n_{Fe}=0,8mol.$
$⇒m_{ddHCl}=\dfrac{0,8.36,5}{20\%}=146g.$
$⇒m_{dd\ spư}=m_{Fe}+m_{ddHCl}-m_{H_2}$
$⇒m_{dd\ spư}=22,4+146-0,4.2=167,6g.$
$⇒C\%_{FeCl_2}=\dfrac{50,8}{167,6}.100\%=30,31\%$
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