a) \(PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\)
b) \(n_{H_2}=n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ V_{H_2\left(ĐKTC\right)}=0,1.22,4=2,24\left(l\right)\)
c) \(n_{HCl}=2n_{Zn}=2.0,1=0,2\left(mol\right)\\ C_{M_{ddHCl}}=\dfrac{0,2}{0,1}=2\left(M\right)\)
Zn + 2HCl -> ZnCl2 + H2
1 2 1 1
nZn = 6,5/65 = 0,1 mol => nH2 = 0,1 mol
VH2 = 0,1 x 22,4 = 2,24 lit
CMHCl = 0,2/0,1 = 2M