a,
$Zn+2HCl\to ZnCl_2+H_2$
b,
$n_{Zn}=\dfrac{3,25}{65}=0,05 mol$
$\Rightarrow n_{H_2}=0,05 mol$
$\to V_{H_2}=0,05.22,4=1,12l$
$n_{ZnCl_2}=0,05 mol$
$\to m_{ZnCl_2}=0,05.136=6,8g$
c,
$n_{HCl}=2n_{H_2}=0,1 mol$
$V_{HCl}=\dfrac{200}{1,025}=195,12ml=0,19512l$
$\to C_{M_{HCl}}=0,1:0,19512=0,5M$
d,
$C\%_{HCl}=\dfrac{0,1.36,5.100}{200}=1,825\%$