$n_{ZnO}=\dfrac{8,1}{81}=0,1mol$
$n_{H_2SO_4}=0,58.4=2,32mol$
$PTHH :$
$ZnO + H_2SO_4\to ZnSO_4+H_2↑$
Theo pt : 1 mol 1 mol
Theo đbài : 0,1 mol 2,32 mol
Tỉ lệ : $\dfrac{0,1}{1}<\dfrac{2,32}{1}$
⇒Saupharnn ứng H2SO4 dư
Theo pt :
$n_{H_2SO_4\ pư}=n_{ZnSO_4}=n_{ZnO}=0,1mol$
$⇒m_{H_2SO_4\ pư}=0,1.98=9,8(g)$
$m_{ZnSO_4}=0,1.161=16,1(g)$
$b.⇒C_{M_{ZnSO_4}}=\dfrac{0,1}{0,58}=0,17M$