a, Ta có pt : \(Zn+2HCL->ZnCl_2+H_2\)
b, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
Theo pt (ở câu a ) , có : \(n_{Zn}=n_{H_2}=0,1\left(mol\right)\)
\(V_{H_2\left(DKTC\right)}=0,1\times22,4=2,24\left(l\right)\)
nZn= \(\dfrac{6,5}{65}\) = 0,1 (mol)
a. PTHH: Zn + 2HCl -> ZnCl2 + H2
0,1 -> 0,1 -> 0,1 (mol)
b. nH2 = 0,1 x 22,4 = 2,24 (l)