$a,PTPƯ:Fe+2HCl\xrightarrow{} FeCl_2+H_2↑$
$n_{Fe}=\dfrac{5,6}{56}=0,1mol.$
$Theo$ $pt:$ $n_{H_2}=n_{Fe}=0,1mol.$
$⇒V_{H_2}=0,1.22,4=2,24l.$
$b,Theo$ $pt:$ $n_{HCl}=2n_{Fe}=0,2mol.$
$⇒m_{HCl}=0,2.36,5=7,3g.$
$⇒m_{ddHCl}=\dfrac{7,3}{20\%}=36,5g.$
$c,m_{H_2}=0,1.2=0,2g.$
$⇒m_{dd spư}=m_{Fe}+m_{ddHCl}-m_{H_2}$
$⇒m_{dd spư}=5,6+36,5-0,2=41,9g.$
$Theo$ $pt:$ $n_{FeCl_2}=n_{Fe}=0,1mol.$
$⇒m_{FeCl_2}=0,1.127=12,7g.$
$⇒C\%_{FeCl_2}=\dfrac{12,7}{41,9}.100\%=30,31\%$
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