4 KO2(s) + 2 H2O(l)  4 KOH(s) + 3 O2(g)

Question

If reaction vessel contains 0.25 mol KO2 and 0.15 mol of H2O what is limiting reactant? How
many moles of oxygen can be produced?

Answer 1

the equation tells you that you will need twice as many moles of KO2 as H2O.
So figure out how many moles of each you have, and see which will run out first. That will limit the reaction.
You will get either
3/4 as many moles of O2 as KO2
or
3/2 as many moles of O2 as H2O

Answer 2

Two separate ways to do this.
4 KO2(s) + 2 H2O(l)  4 KOH(s) + 3 O2(g)
0.25 mol..... 0.15 mol
1. short way. Choose either, let's pick KO2. Convert 0.25 mols to water needed.
0.25 mols KO2 x (2 mols H2O/4 mols KO2) = 0.125 mols H2O needed. Do you have that much H2O. Yes, therefore, KO2 is the limiting reagent (LR) and H2O is the excess reagent (ER). Or choose the other one. Convert 0.15 moles H2O to moles KO2 needed.
0.15 moles H2O x (4 mols KO2/2 moles H2O) = 0.30 moles KO2 needed. Do you have that much KO2. No, you have only 0.25 moles KO2; therefore, KO2 is the LR.
2. the long way. Pick a product II'll pick KOH) and calculate the moles possible for each reactant.
First, KO2. 0.25 mols KO2 x (4 moles KOH/4 mol KO2) = 0.25 mols KOH produced.
Next, H2O. 0.15 moles H2O x (4 mols KOH/2 mols H2O) = 0.30 mol KOH produced.
The LR is ALWAYS the smaller of the two; therefore, KO2 is the LR and H2O is the ER.