\(n_{CuO}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\underrightarrow{BTNT.Cu}n_{Cu}+n_{CuOdư}=0,15\\m_{\text{chất rắn}}=64n_{Cu}+80n_{CuOdư}=10\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Cu}=0,125\left(mol\right)\\n_{CuOdư}=0,025\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{CuOpư}=0,125\left(mol\right)\)
\(\Rightarrow H=\dfrac{0,125}{0,15}.100\%=83,33\%\)