$a,PTPƯ:Fe+H_2SO_4\xrightarrow{} FeSO_4+H_2↑$
$b,n_{Fe}=\dfrac{2,24}{56}=0,04mol.$
$Theo$ $pt:$ $n_{Fe_SO_4}=n_{Fe}=0,04mol.$
$⇒m_{FeSO_4}=0,04.152=6,08g.$
$Theo$ $pt:$ n_{H_2}=n_{Fe}=0,04mol.$
$c,PTPƯ:CuO+H_2\xrightarrow{t^o} Cu+H_2O$
$Theo$ $pt:$ $n_{Cu}=n_{H_2}=0,04mol.$
$⇒m_{Cu}=0,04.64=2,56g.$
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