mchất rắn không tan = mCu = 3,5 (g)
\(n_{NaOH}=0,05.1=0,05\left(mol\right)\)
PTHH:
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+2NaCl\)
Theo PT: \(n_{CuO}=n_{CuCl_2}=\dfrac{1}{2}n_{NaOH}=0,025\left(mol\right)\)
=> \(\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{3,5}{0,025.80+3,5}.100\%=63,64\%\\\%m_{CuO}=100\%-63,64\%=36,36\%\end{matrix}\right.\)