\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\
pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,2 0,2 0,2
\(m_{ZnCl_2}=0,2.136=40,8g\\
n_{CuO}=\dfrac{24}{60}=0,3\left(mol\right)\\
pthh:CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
\(LTL:\dfrac{0,3}{1}>\dfrac{0,2}{1}\)
=> CuO dư , H2 Hết
\(n_{CuO\left(p\text{ư}\right)}=n_{Cu}=n_{H_2}=0,2\left(mol\right)\\
\left\{{}\begin{matrix}m_{Cu}=0,2.64=12,8g\\m_{CuO\left(d\right)}=\left(0,3-0,2\right).80=8g\end{matrix}\right.\)
\(m_{CR}=12,8+8=20,8g\)