a) PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b) Ta có: \(n_{H_2}=\frac{2,24}{22,4}=0,1\left(mol\right)=n_{Zn}\)
\(\Rightarrow m_{Zn}=65\cdot0,1=6,5\left(g\right)\) \(\Rightarrow m_{Cu}=3,5\left(g\right)\)
c) Theo PTHH: \(n_{HCl}=2n_{Zn}=0,2mol\)
\(\Rightarrow V_{HCl}=\frac{0,2}{0,2}=1\left(l\right)\)