Question

Potassium superoxide, KO2, reacts with carbon dioxide to form potassium carbonate and oxygen:

4KO₂(s) + 2 CO₂ --> 2K₂CO₃(s) + 3O₂(g)
This reaction makes potassium superoxide useful in a self-contained breathing apparatus. How much O₂ could be produced from 2.5 grams of KO2 and 4.5 grams of CO₂?

Answer 1

To find how much O₂ is produced from the reaction of KO₂ with CO₂, write the balanced chemical equation, calculate the molar mass of the reactants, determine the limiting reactant, and convert this to the mass of O₂ using stoichiometry.

Step-by-step explanation:

To calculate the amount of O₂ that could be produced from 2.5 grams of KO₂ and 4.5 grams of CO₂, we need to follow the stoichiometry of the balanced chemical equation.

Step 1: Write the balanced equation.

4KO₂(s) + 2CO₂(g) → 2K₂CO₃(s) + 3O₂(g)

Step 2: Calculate the molar mass of KO₂ and O₂.

Molar mass of KO₂ = 39.1 (K) + 32 (O₂) = 71.1 g/mol

Molar mass of O₂ = 32 g/mol

Step 3: Convert grams to moles.

Moles of KO₂ = 2.5 g / 71.1 g/mol

Moles of CO₂ = 4.5 g / 44.01 g/mol (Molar mass of CO₂)

Step 4: Using stoichiometry, find the limiting reactant.

According to the balanced equation, 4 moles of KO₂ produce 3 moles of O₂.

Calculate the theoretical yield of O₂ based on KO₂ and CO₂ separately and identify which one produces less O₂. This is the limiting reactant.

Step 5: Calculate the amount of O₂ produced.

Use the moles of the limiting reactant and the stoichiometry of the reaction to find how many moles of O₂ can be produced.

Step 6: Convert moles of O₂ to grams.

Since O₂ is the product, use the molar mass of O₂ to get the mass from the moles of O₂.

By going through these steps, one can determine how much O₂ could be produced in the reaction.