\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(n_{Cu}=\dfrac{40}{80}=0,5\left(mol\right)\)
PTHH: \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\left(1\right)\)
Xét tỉ lệ: \(\dfrac{0,5}{1}>\dfrac{0,1}{1}\) => CuO dư, tính theo H2
Theo PT \(\left(1\right):n_{Cu}=n_{H_2}=0,1\left(mol\right)\)
\(\rightarrow m_{Cu}=0,1.64=6,4\left(g\right)\)