How many moles of KCLO3 are needed to produce 50 moles of O2?

You'd need 33.3 moles of potassium chlorate, ##KClO_3##, to produce that much oxygen.

All you need in order to answer this question is the balanced chemical equation for the of potassium chlorate

##color(blue)(2)KClO_(3(s)) -> 2KCl_((s)) + color(red)(3)O_(2(g))##

Notice that you have a ##color(blue)(2):color(red)(3)## between potassium chlorate and oxygen gas, which means that, regardless of how many moles of the former react, you'll always produce 3/2 times more moles of the latter.

Since you know how many moles of oxygen you need the reaction to produce, you can work backwards and determine how many moles of potassium chlorate you need by using the same mole ratio

##50cancel("moles" O_2) * (color(blue)(2)" moles "KClO_3)/(color(red)(3)cancel("moles "O_2)) = "33.3 moles "## ##KClO_3##

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