$a,PTPƯ:Fe+2HCl\xrightarrow{} FeCl_2+H_2↑$
$b,n_{Fe}=\dfrac{42}{56}=0,75mol.$
$Theo$ $pt:$ $n_{HCl}=n_{Fe}=1,5mol.$
$⇒m_{HCl}=1,5.36,5=54,75g.$
$⇒m_{ddHCl}=\dfrac{54,75}{25\%}=219g.$
$c,Theo$ $pt:$ $n_{H_2}=n_{Fe}=0,75mol.$
$⇒V_{H_2}=0,75.22,4=16,8l.$
$d,Theo$ $pt:$ $n_{FeCl_2}=n_{Fe}=0,75mol.$
$⇒m_{FeCl_2}=0,75.127=95,25g.$
$⇒m_{ddFeCl_2}=m_{Fe}+m_{ddHCl}-m_{H_2}=42+219-(0,75.2)=259,5g.$
$⇒C\%_{FeCl_2}=\dfrac{95,25}{259,5}.100\%=36,7\%$
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