$n_{Zn}$ = $\frac{m_{Zn}}{M_{Zn}}$ = $\frac{13}{65}=0,2(mol)$
$Zn_{}$ + $2HCl_{}$ → $ZnCl_{2}$ + $H_{2}$
0,2 → 0,4 0,2 0,2 : mol
a) $V_{H_2}$ = $n_{H_2}.22,4=0,2.22,4=4,48(l)$
b) $m_{HCl}$ = $n_{HCl}$ . $M_{HCl}=0,4.36,5=14,6(g)$
C%=$\frac{m_{HCl}}{m_{dung dịchHCl}}$.100% = $\frac{14,6}{80}$.100%=18,25%
c) $m_{dung dịch SPƯ}$ = $(m_{Zn}$ + $m_{dung dịch HCl})$ - $m_{H_2}$
$m_{dung dịch SPƯ} =(13+80)-0,2.2=92,6(g)$
$m_{ZnCl_2}$ = $n_{ZnCl_2}$ . $M_{ZnCl_2}=0,2.136=27,2(g)$
C%=$\frac{m_{ZnCl_2}}{m_{dung dịch SPƯ}}$ .100%= $\frac{27,2}{92,6}$.100% ≈ 29,37%