$n_{Zn}$ = $\frac{m_{Zn}}{M_{Zn}}$ = $\frac{19,5}{65}=0,3(mol)$
$m_{HCl}$= $\frac{m_{dung dichjHCl}.C}{100}$ = $\frac{262,8.10}{100}=26,28(g)$
$n_{HCl}$ = $\frac{m_{HCl}}{M_{HCl}}$ = $\frac{26,28}{36,5}=0,72(mol)$
$Zn_{}$ + $2HCl_{}$ → $ZnCl_{2}$ + $H_{2}$
0,3 0,72 :TPƯ
0,3 0,6 :PƯ
0 0,12 :SPƯ
$\frac{0,3}{1}$ < $\frac{0,72}{2}$=>$HCl_{}$ dư, $Zn_{}$ hết => Tính theo $Zn_{}$
$m_{HCl dư}$ = $n_{HCl dư}$ . $M_{HCl}=0,12.36,5=4,38(g)$