How many mL of 1.2M potassium superoxide (KO2) are consumed in a reaction with water and...?

How many mL of 1.2M potassium superoxide (KO2) are consumed in a reaction with water and carbon dioxide if 210 mL of 0.142 M potassium carbonate (KHCO3) are produced?
4KO2 + 2H2O + 4CO2 --> 4KHCO3 + 3O2

Answer:
Look at the chemical equation.
4 moles of superoxide gives 4 moles of carbonate. So, each mole of superoxide gives one mole of carbonate.

Number of moles of carbonate formed = 0.210*0.142=2.982*10^-2 moles

So, the number of moles of superoxide = 2.982*10^-2
This is equal to molarity*volume of solution.
So, 1.2*Volume = 2.982*10^-2
Volume = 2.485*10^-2 L = 24.85 mL of KO2 is consumed.

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