a, $Zn+2HCl\to ZnCl_2+H_2$

b, 

$n_{Zn}=\frac{13}{65}=0,2 mol$

$n_{HCl}=\frac{100.14,6\%}{36,5}=0,4 mol$

$\Rightarrow n_{H_2}= n_{ZnCl_2}=0,2 mol$ (phản ứng vừa đủ)

$V_{H_2}=0,2.22,4=4,48l$

c,

$m_{dd spứ}=13+100-0,2.2=112,6g$

$\Rightarrow C\%_{ZnCl_2}=\frac{0,2.136.100}{112,6}= 24,16\%$

$n_{Zn}=13/65=0,2mol$

$n_{HCl}=100.14,6\%=14,6g⇒n_{HCl}=14,6/36,5=0,4mol$

$a/Zn+2HCl\to ZnCl_2+H_2↑$

$\text{⇒Sau pư cả 2  chất đều hết.}$

$\text{b/Theo pt :}$

$n_{H_2}=n_{Zn}=0,2mol$

$⇒V_{H_2}=0,2.22,4=4,48l$

$\text{c/Theo pt :}$

$n_{ZnCl_2}=n_{Zn}=0,2mol$

$⇒m_{ZnCl_2}=0,2.136=27,2g$

$m_{dd spu}=13+100-0,2.2=112,6g$

$⇒C\%_{ZnCl_2}=\dfrac{27,2}{112,6}.100\%=24,16\%$