a) \(n_{Zn}=\dfrac{3,25}{65}=0,05\left(mol\right)\)
PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
0,05--->0,05------>0,05-->0,05
=> \(V_{H_2}=0,05.22,4=1,12\left(l\right)\)
b) \(m_{H_2SO_4}=0,05.98=4,9\left(g\right)\Rightarrow m_{dd.H_2SO_4}=\dfrac{4,9.100}{20}=24,5\left(g\right)\)
=> mdd sau pư = 3,25 + 24,5 - 0,05.2 = 27,65 (g)
\(C\%_{ZnSO_4}=\dfrac{0,05.161}{27,65}.100\%=29,114\%\)