Dude pls help

4KO2 + 2H2O → 4KOH + 3O2

If a reaction vessel contains 0.15 mol KO2 (superperoxide)and 0.10 mol H2O, what is the limiting reactant?

How many moles of oxygen will be produced?

KO is the limiting reactant.

0.11 mol O₂ will be produced.

Explanation:

4 KO₂  +  2 H₂O  ⇒  4 KOH  +  3 O₂

Find the limiting reagent by dividing the moles of the reactant by the coefficient in the equation.

(0.15 mol KO₂)/4 = 0.0375

(0.10 mol H₂O)/2 = 0.05

KO₂ is the limiting reagent.

The amount of product produced depends on the limiting reagent.  To find how much is produced, take moles of limiting reagent and multiply it by the ratio of reagent to product.  You can find the ratio by looking at the equation.  For every 4 moles of KO₂, 3 moles of O₂ are produced.

0.15 mol KO₂  (3 mol O₂)/(4 mol KO₂) = 0.1125 mol O₂

0.11 mol O are produced.

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the answer is 1 i believe