Suppose there is 100 g of base, in which 0.65*100 = 65 g KOH is contained.
It corresponds to 65 g / 56.1 g/mole = 1.158 mole of KOH, so that the number of moles of K2O is 1.158/2 = 0.579 mole of K2O.
Mass of K2O is 0.579 mole * 94.2 g/mole = 54.57 g.
Dividing by 100 g of the base, the percentage of K2O is 54.57 %.
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