\(n_{K_2O}=\dfrac{28,2}{94}=0,3\left(mol\right)\\ n_{H_2O}=\dfrac{40}{18}=\dfrac{20}{9}\left(mol\right)\)
PTHH: K2O + H2O ---> 2KOH
LTL: \(0,3< \dfrac{20}{9}\) => H2O dư
Thep pthh: \(n_{KOH}=2n_{K_2O}=2.0,3=0,6\left(mol\right)\)
=> mKOH = 0,6.56 = 33,6 (g)