a) $Zn + 2HCl \to ZnCl_2 + H_2$
b) $n_{Zn} = \dfrac{0,65}{65} = 0,01(mol)$
Theo PTHH :
$n_{HCl} = 2n_{Zn} = 0,02(mol) \Rightarrow m_{HCl} = 0,02.36,5 = 0,73(gam)$
$n_{ZnCl_2} = n_{H_2} = n_{Zn} = 0,02(mol)$
Suy ra :
$m_{ZnCl_2} = 0,02.136 = 2,72(gam)$
$V_{H_2} = 0,02.22,4 = 0,448(lít)$
