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Question

Potassium superoxide, KO₂, reacts with carbon dioxide to form potassium carbonate and oxygen: 4KO₂ + 2CO₂ —> 2K₂CO₃ + 3O₂

This reaction makes potassium superoxide useful in a self-contained breathing apparatus. How much O₂ could be produced from 2.49 g of KO₂ and 4.44 g of CO₂?

___ g

Answer 1

Hello, Kirsten,

I had answered this earlier, but don't see it. I'll repeat it, but there may be another one showing up.

We need to convert everything into moles. The molar masses, in g/mole:

KO2 71.1

CO2 44.0

O2 32

Moles KO2 = 2.49g/(71.1g/mole) = 0.035 moles

Moles CO2 = 4.44g/(44g/mole) = 0.10 moles

4KO₂ + 2CO₂ —> 2K₂CO₃ + 3O₂

The reaction requires twice as many moles of KO2 than moles of CO2. That means 0.10 mole of CO2 would consume 0.20 moles of KO2. But we only have 0.035 moles of KO2, so the limiting reagent is KO2.

There are 3 moles of O2 produced for every 2 moles of CO2 consumed. In this case we'll assume all of the KO2 reacts, so that should result in (3/4) * (0.035 moles) = 0.026 moles of O2.

To get the mass, 0.026 mole * (32g/mole) = 0.841 grams O2

Not much, but we only had 2.49g of KO2 to start with.

Breath gently.

Bob

Answer 2

This is a limiting reagent problem. the way to figure it out is to do the stoichiometry for each amount given, calculating the amount of Oxygen produced. The smallest amount is the answer because that reactant "ran out".

for KO₂

2.49g KO₂ x 1 mole KO₂ x 3 mole O₂ x 32.00 g O₂ = 0.841 g O₂

71.10 g KO₂ 4 mole KO₂ 1 mole O₂

for CO₂

4.44g CO₂ x 1 mole CO₂ x 3 mole O₂ x 32.00g O₂ = 4.84 g O₂

44.01 g CO₂ 2 mole CO₂ 1 mole O₂

Therefore the answer is 0.841g O₂ since it is the smallest amount produced meaning KO₂ is the limiting reagent.