$n_{Zn}=13/65=0,2mol$
$n_{HCl}=100.14,6\%=14,6g⇒n_{HCl}=14,6/36,5=0,4mol$
$a/Zn+2HCl\to ZnCl_2+H_2↑$
$\text{⇒Sau pư cả 2 chất đều hết.}$
$\text{b/Theo pt :}$
$n_{H_2}=n_{Zn}=0,2mol$
$⇒V_{H_2}=0,2.22,4=4,48l$
$\text{c/Theo pt :}$
$n_{ZnCl_2}=n_{Zn}=0,2mol$
$⇒m_{ZnCl_2}=0,2.136=27,2g$
$m_{dd spu}=13+100-0,2.2=112,6g$
$⇒C\%_{ZnCl_2}=\dfrac{27,2}{112,6}.100\%=24,16\%$
