I think you means the KO2 reacts with H2O. The equation of this reaction is 4KO2+2H2O->4KOH +3O2. The ratio of mole number of O2 and KO2 is 3:4. So the mole number of O2 produced is 0.500/4*3=0.375 mol.
<h2>Answer:</h2>

<h2>Explanations</h2>
The complete balanced equation for the given reaction is expressed as;

Given the following parameters
Mass of CH4 = 5.90×10^−3 g = 0.0059grams
Determine the moles of methane

According to stoichimetry, 1 mole of methane produces 2 moles of water, hence the moles of water required will be:

Determine the mass of water produced

Therefore the mass of water produced from the complete combustion of 5.90×10−3 g of methane is 1.33 * 10^-2grams
Answer:
Rubidium is a typical but very reactive member of the series of alkali metals. It is appreciably more reactive than potassium, but less so than caesium, and so would be expected to react more violently with those materials that are hazardous with potassium or sodium.