$n_{ZnO}=\dfrac{8,1}{81}=0,1mol \\n_{HCl}=0,2.0,2=0,04mol \\a.PTHH :$
$ZnO + 2HCl\to ZnCl_2+H_2O$
$\text{b.Theo pt : 1 mol 2 mol}$
$\text{Theo đbài : 0,1 mol 0,04 mol}$
Tỷ lệ : $\dfrac{0,1}{1}>\dfrac{0,04}{2}$
$\text{⇒Sau pư ZnO dư}$
$\text{Theo pt :}$
$n_{ZnO\ pư}=\dfrac{1}{2}.n_{HCl}=\dfrac{1}{2}.0,04=0,02mol \\⇒n_{ZnO\ dư}=0,1-0,02=0,08mol \\n_{ZnCl_2}=\dfrac{1}{2}.n_{HCl}=\dfrac{1}{2}.0,04=0,02mol \\⇒C_{M_{ZnO\ dư}}=\dfrac{0,08}{0,2}=0,4M \\C_{M_{ZnCl_2}}=\dfrac{0,02}{0,2}=0,1M$