$a,PTPƯ:Fe+2HCl\xrightarrow{} FeCl_2+H_2↑$
$n_{Fe}=\dfrac{5,6}{56}=0,1mol.$
$Theo$ $pt:$ $n_{H_2}=n_{Fe}=0,1mol.$
$⇒V_{H_2}=0,1.22,4=2,24l.$
$b,Theo$ $pt:$ $n_{HCl}=2n_{Fe}=0,2mol.$
$⇒m_{HCl}=0,2.36,5=7,3g.$
$⇒m_{ddHCl}=\dfrac{7,3}{7,3\%}=100g.$
$c,Theo$ $pt:$ $n_{FeCl_2}=n_{Fe}=0,1mol.$
$⇒C\%_{FeCl_2}=\dfrac{0,1.127}{5,6+100-(0,1.2)}.100\%=12,05\%$
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