\(\begin{array} {l} a)\\ Zn+H_2SO_4\to ZnSO_4+H_2\\ b)\\ n_{Zn}=\dfrac{32,5}{65}=0,5(mol)\\ n_{H_2SO_4}=\dfrac{100.39,2\%}{98}=0,4(mol)\\ \text{Vì }n_{H_2SO_4}<n_{Zn}\to Zn\text{ dư}\\ \text{Theo PT: }n_{H_2}=n_{H_2SO_4}=0,4(mol)\\ \to V_{H_2}=0,4.22,4=8,96(l)\\ c)\\ \text{Theo PT: }n_{ZnSO_4}=n_{H_2SO_4}=0,4(mol)\\ \to m=m_{ZnSO_4}=0,4.161=64,4(g) \end{array}\)
\(n_{Zn}=\dfrac{32,6}{65}=0,5\left(mol\right)\\ n_{H_2SO_4}=\dfrac{100.39,2\%}{98}=0,4\left(mol\right)\\ pthh:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)
\(\dfrac{0,5}{1}>\dfrac{0,4}{1}\)
=>Zn dư
\(n_{H_2}=n_{H_2SO_4}=0,4\left(mol\right)\\ V_{H_2}=0,4.22,4=8,96\left(l\right)\)
\(n_{ZnSO_4}=n_{H_2SO_4}=0,4\left(mol\right)\\
m_{ZnSO_4}=0,4.161=64,4\left(g\right)\)