B1 :
PTHH : Zn + 2HCl ->(t*) ZnCl2 + H2
Theo đề bài ta có : nZn = \(\dfrac{m}{M}=\dfrac{6.5}{65}=0,1\left(mol\right)\)
Theo PTHH ta có \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\)
=> \(V_{H_2\left(\text{đ}ktc\right)}=n.22,4=0,1.22,4=2,24\left(l\right)\)
Theo PTHH ta có \(n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\)
=> \(m_{ZnCl_2}=n.M=0,1.136=13,6\left(g\right)\)