Ta có: \(n_{ZnO}=\dfrac{16,2}{81}=0,2\left(mol\right)\)
\(m_{H_2SO_4}=250.19,6\%=49\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\)
PT: \(ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)
Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,5}{1}\), ta được H2SO4 dư.
Theo PT: \(n_{ZnSO_4}=n_{ZnO}=0,2\left(mol\right)\)
\(\Rightarrow m_{ZnSO_4}=0,2.161=32,2\left(g\right)\)
\(PTPU:ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)
\(0,2:0,2:0,2:0,2\left(mol\right)\)
\(n_{ZnO}=\dfrac{m}{M}=\dfrac{16,2}{81}=0,2\left(mol\right)\)
\(m_{ZnSO_4}=n.M=0,2.161=32,2\left(g\right)\)
