\(Zn\left(0,1\right)+2HCl\left(0,2\right)\rightarrow ZnCl_2+H_2\left(0,1\right)\)
\(V_{ddHCl}=\dfrac{0,2}{2}=0,1M\)
\(H_2+CuO\rightarrow Cu+H_2O\)
\(n_{CuO}=\dfrac{24}{80}=0,3\left(mol\right)\)
Sau pư: H2 hết, CuO dư
\(\Rightarrow m_{Cu}=0,1.64=6,4g.\)