a. \(n_{ZnO}=\dfrac{8.1}{81}=0,1\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{100.19,6\%}{98}=0,2\left(mol\right)\)
PTHH : ZnO + H2SO4 -> ZnSO4 + H2O
0,1 0,1 0,1 0,1
Ta thấy : \(\dfrac{0.1}{1}< \dfrac{0.2}{1}\Rightarrow ZnO.đủ,H_2SO_4.dư\)
\(m_{ZnSO_4}=0,1.161=16,1\left(g\right)\)
b. \(C\%_{H_2SO_4.dư}=\dfrac{\left(0,2-0,1\right).98}{8,1+100}.100\approx9,1\%\)