\(n_{CuO}=\dfrac{16}{80}=0,2mol\\
n_{HCl}=0,25.2=0,5mol\\
CuO+2HCl\rightarrow CuCl_2+H_2O\\
\Rightarrow\dfrac{0,2}{1}< \dfrac{0,5}{2}\Rightarrow HCl.dư\\
CuO+2HCl\rightarrow CuCl_2+H_2O\)
0,2 0,4 0,2
\(a)m_{ddHCl}=250.1,2=300g\\
m_{dd.sau.pứ}=16+300=316g\\
C_{\%CuCl_2}=\dfrac{0,2.135}{316}\cdot100\%=8,54\%\\
C_{\%HCl.dư}=\dfrac{\left(0,5-0,4\right)36,5}{316}\cdot100\%=1,16\%\\
b)C_{M_{CuCl_2}}=\dfrac{0,2}{0,25}=0,8M\\
C_{M_{HCl.dư}}=\dfrac{0,5-0,4}{0,25}=0,4M\)