`a.`

PTHH : `Zn+2HCl->ZnCl_2+H_2`

`b.`

`n_{Zn}=(13)/(65)=0,2` `mol`

Theo phương trình : $\begin{cases} n_{HCl}=2n_{Zn}=0,4(mol)\\n_{ZnCl_2}=n_{Zn}=0,2(mol)\\n_{H_2}=n_{Zn}=0,2(mol) \end{cases}$

`=>` $\begin{cases} m_{HCl(dd)}=0,4.36,5:\dfrac{10}{100}=146(g)\\m_{ZnCl_2}=0,2.136=27,2(g)\\m_{H_2}=0,2.2=0,4(g) \end{cases}$

`=>C%ZnCl_2=(27,2)/(146+13-0,4).100%~~17,15%`