In a self-contained breathing apparatus, potassium superoxide, KO₂, reacts with carbon dioxide to form potassium carbonate and oxygen. Identify the limiting reactant and determine the amount of oxygen gas, in moles, that can be produced from 250 g of KO₂ and 450 g of CO₂.
4KO₂(s) + 2CO₂(g) $\rightarrow$ 2K₂CO₃(s) + 3O₂(g)

Question

In a self-contained breathing apparatus, potassium superoxide, KO₂, reacts with carbon dioxide to form potassium carbonate and oxygen. Identify the limiting reactant and determine the amount of oxygen gas, in moles, that can be produced from 250 g of KO₂ and 450 g of CO₂.
4KO₂(s) + 2CO₂(g)

\rightarrow

2K₂CO₃(s) + 3O₂(g)

Quizplus · Accepted Answer

To determine the limiting reactant, we need to use stoichiometry and compare the amounts of products that can be formed from each reactant. First, we need to convert the masses of KO₂ and CO₂ to moles: - Moles of KO₂: 250 g / 71.10 g/mol = 3.52 mol - Moles of CO₂: 450 g / 44.01 g/mol = 10.22 mol Next, we can use the balanced chemical equation to calculate the theoretical amount of oxygen gas that can be produced from each reactant: - From KO₂: 3.52 mol KO₂ x (3 mol O2 / 4 mol KO₂) = 2.64 mol O2 - From CO₂: 10.22 mol CO₂ x (3 mol O2 / 2 mol CO₂) = 15.33 mol O2 Therefore, KO₂ is the limiting reactant, and the amount of oxygen gas that can be produced is 2.64 mol. The correct choice is C.

In a Self-Contained Breathing Apparatus, Potassium Superoxide, KO2, Reacts with Carbon

In a Self-Contained Breathing Apparatus, Potassium Superoxide, KO₂, Reacts with Carbon