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

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:
Moles KHCO3 = Molarity x V = 0.142 moles/L x 0.210 L =
= 0.0298

the ratio between KO2 and KHCO3 is 4 : 4 so moles KO2 = 0.0298

Molarity = moles / L

L = moles / Molarity = 0.0298 / 1.2 = 0.0248 L = 24.8 mL

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