\(a)CTHH:ZnSO_4\\ b)n_{Zn}=\dfrac{0,65}{65}=0,01mol\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
0,01 0,01
\(V_{H_2\left(đktc\right)}=0,01.22,4=0,224l\\ V_{H_2\left(đkc\right)}=0,01.24,79=0,2479l\)
\(n_{Zn}=\dfrac{0,65}{65}=0,01\left(mol\right)\\ a,Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ b,n_{H_2}=n_{Zn}=0,01\left(mol\right)\\ Vậy:V_{H_2\left(đktc\right)}=0,01.22,4=0,224\left(l\right)\\ V_{H_2\left(đkc\right)}=0,01.24,79=0,2479\left(l\right)\)