a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, Ta có: \(n_{HCl}=\dfrac{49.15\%}{36,5}\approx0,2\left(mol\right)\)
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,1.24,79=2,479\left(l\right)\)
c, \(n_{Zn}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\)
\(\Rightarrow m_{Zn}=0,1.65=6,5\left(g\right)\)