a/ $Zn+H_2SO_4\to ZnSO_4+H_2$
b/ $n_{Zn}=\dfrac{13}{65}=0,2(mol)$
$n_{H_2SO_4}=5.0,1=0,5(mol)$
Xét tỉ lệ: $\dfrac{n_{Zn}}{1}<\dfrac{n_{H_2SO_4}}{1}(0,2<0,5mol)$
$→H_2SO_4$ dư $0,3mol$
$→m_{H_2SO_4\,\text{dư}}=0,3.98=29,4g$
Theo phương trình: $n_{ZnSO_4}=n_{H_2}=n_{Zn}$
$→n_{ZnSO_4}=n_{H_2}=0,2(mol)\\→\begin{cases}m_{ZnSO_4}=0,2.161=32,2g\\m_{H_2}=0,2.2=0,4g\end{cases}$
Vậy $m_{H_2SO_4\,\text{dư}}=29,4g;\,m_{ZnSO_4}=32,2g;\,m_{H_2}=0,4g$
c/ $V_{H_2}=0,2.22,4=4,48l$
Vậy $V_{H_2}=4,48l$