$PTPƯ:CuO+H_2\xrightarrow{t^o} Cu+H_2O$
$n_{H_2}=\dfrac{3,36}{22,4}=0,15mol.$
$n_{CuO}=\dfrac{24}{80}=0,3mol.$
$\text{Lập tỉ lệ:}$ $\dfrac{0,3}{1}>\dfrac{0,15}{1}$
$⇒CuO$ $dư.$
$⇒n_{CuO}(dư)=0,3-\dfrac{0,15.1}{1}=0,15mol.$
$⇒m_{CuO}(dư)=0,15.80=12g.$
$b,Theo$ $pt:$ $n_{Cu}=n_{H_2O}=n_{H_2}=0,15mol.$
$⇒m_{Cu}=0,15.64=9,6g.$
$⇒m_{H_2O}=0,15.18=2,7g.$
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