a) Zn + H2SO4 $\to$ ZnSO4 + H2
b) n H2 = n Zn = 0,85/65 = 17/1300 mol
=> V H2 = 17/1300 .22,4 = 0,292(lít)
PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)
Ta có: \(n_{Zn}=\dfrac{0,65}{65}=0,01\left(mol\right)=n_{H_2}\) \(\Rightarrow V_{H_2}=0,01\cdot22,4=0,224\left(l\right)\)
*P/s: Sửa 0,85 \(\rightarrow\) 0,65