$a)$
$ Zn + 2HCl \to ZnCl_2 + H_2 \uparrow $
$b)$
$n_{Zn} = \dfrac{16,25}{65}=0,25\ mol$
Theo PT ta có $n_{HCl} = 2 n_{Zn} = 2. 0,25 = 0,5\ mol$
$\to m_{HCl} = 0,5 . 36, 5=18,25\ g$
$\to C\% = \dfrac{18,25}{500} . 100 \% =3,65 \%$
$c)$
$ m_{H_2O} = m_{ddHCl} - m_{HCl} = 500 - 18,25 = 481,75\ g$
Theo PT ta có $n_{ZnCl_2} = n_{Zn} = 0,25\ mol$
$\to m_{ZnCl_2} = 0,25 .136 = 34\ g$
$ \to m_{ddZnCl_2} = 34 + 481,75 =515,75\ g$
$\to C\% = \dfrac{34}{515,75} . 100 \% \approx 6,59 \%$