Ta có
a. PTHH
\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
b. Có \(n_{Zn}=\frac{6,5}{65}=0,1\left(mol\right)\)
Theo PTHH ta có
\(n_{HCl}=2n_{Zn}=0,2\left(mol\right)\Rightarrow V_{HCl}=\frac{0,2}{1}=0,2\left(mol\right)\)
c. PTHH
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)
Có \(n_{H_2SO_4}=n_{Zn}=0,1\left(mol\right)\Rightarrow m_{H_2SO_4}=0,1\times98=9,8\left(mol\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\frac{9,8}{49}.100=20\left(g\right)\)