\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Ta có :
\(n_{H2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)
\(n_{Zn}=n_{H2}=0,2\left(mol\right)\)
\(\%m_{Zn}=\frac{0,2.65}{20}.100\%=65\%\)
\(\%m_{Cu}=100\%-65\%=35\%\)
\(n_{HCl}=2n_{H_2}=0,4\left(mol\right)\)
\(\rightarrow V_{HCl}=\frac{0,4}{0,5}=0,8\left(l\right)\)