`n_{Cu}=\frac{36,096}{64}=0,564(mol)`
`n_{CuO}=\frac{48}{80}=0,6(mol)`
`CuO+H_2\overset{t^o}{\to}Cu+H_2O`
Do `\frac{0,6}{1}>\frac{0,564}{1}`
`=>` `CuO` dư
`=> n_{CuO \text{pứ}}=n_{Cu}=0,564(mol)`
`=> %H=\frac{m_{CuO \text{pứ}}}{m_{CuO \text{lt}}}.100%=\frac{0,564.80}{48}.100%=94%`