$a,PTPƯ:Fe+2HCl\xrightarrow{} FeCl_2+H_2↑$
$n_{Fe}=\dfrac{5,6}{56}=0,1mol.$
$Theo$ $pt:$ $n_{FeCl_2}=n_{Fe}=0,1mol.$
$⇒m_{FeCl_2}=0,1.127=12,7g.$
$b,Theo$ $pt:$ $n_{H_2}=n_{Fe}=0,1mol$
$⇒V_{H_2}=0,1.22,4=2,24l.$
$c,PTPƯ:CuO+H_2\xrightarrow{t^o} Cu+H_2O$
$n_{CuO}=\dfrac{6,4}{80}=0,08mol.$
$\text{Lập tỉ lệ:}$ $\dfrac{0,08}{1}<\dfrac{0,1}{1}$
$⇒H_2$ 4dư.$
$Theo$ $pt:$ $n_{Cu}=n_{CuO}=0,08mol.$
$⇒m_{Cu}=0,08.64=5,12g.$
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