$n_{Zn}=\dfrac{13}{65}=0,2mol$
$a.Zn+2HCl\to ZnCl_2+H_2↑$
b.Theo pt :
$n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,2mol$
$⇒m_{ZnCl_2}=0,2.136=27,2g$
$V_{H_2}=0,2.22,4=4,48l$
$c.n_{HCl}=2.n_{Zn}=0,2.2=0,4mol$
$⇒m_{HCl}=0,4.36,5=14,6g$
$m_{dd\ spu}=13+250-0,2.2=262,6g$
$⇒C\%_{ZnCl_2}=\dfrac{27,2}{262,6}.100\%=10,36\%$
$C\%_{HCl}=\dfrac{14,6}{250}.100\%=5,84\%$