Question

Answer 1

Ta có: \(n_{ZnO}=\dfrac{16}{81}\left(mol\right)\)

\(a.PTHH:ZnO+H_2SO_4--->ZnSO_4+H_2O\)

b. Theo PT: \(n_{H_2SO_4}=n_{ZnO}=\dfrac{16}{81}\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=\dfrac{16}{81}.98=19,36\left(g\right)\)

Answer 2

a) \(ZnO+H_2SO_4->ZnSO_4+H_2O\)

b) Ta có: \(n_{ZnO}=\dfrac{16}{81}\approx0,2\left(mol\right)\)

=> \(n_{H_2SO_4}=0,2\left(mol\right)\)

=> \(m_{H_2SO_4}=0,2.98=19,6\left(g\right)\\ =>m_{ddH_2SO_4}=\dfrac{19,6x100}{20}=98\left(g\right)\)