A self-contained breathing apparatus contains 25.0 g KO₂ that is used to treat 45.0 g CO₂ to regenerate oxygen. How many grams of CO₂ were not converted to oxygen because of the limited amount of potassium superoxide present?
4KO₂(s) + 2CO₂(g) $\rightarrow$ 2K₂CO₃(s) + 3O₂(g)

Question

A self-contained breathing apparatus contains 25.0 g KO₂ that is used to treat 45.0 g CO₂ to regenerate oxygen. How many grams of CO₂ were not converted to oxygen because of the limited amount of potassium superoxide present?
4KO₂(s) + 2CO₂(g)

\rightarrow

2K₂CO₃(s) + 3O₂(g)

Quizplus · Accepted Answer

First, we need to balance the chemical equation: 4KO₂(s) + 2CO₂(g) -> 2K₂CO₃(s) + 3O₂(g) From the balanced equation, we can see that for every 4 moles of KO₂ used, 2 moles of CO₂ are consumed, and 3 moles of O2 are produced. To find the limiting reactant, we need to calculate how many moles of each reactant are present: n(KO₂) = m(KO₂) / M(KO₂) = 25.0 g / 71.10 g/mol = 0.352 moles n(CO₂) = m(CO₂) / M(CO₂) = 45.0 g / 54.09 g/mol = 0.832 moles Based on the balanced equation, 0.352 moles of KO₂ will react with 0.176 moles of CO₂ to produce 0.264 moles of O2. This means that the remaining 0.656 moles of CO₂ were not converted to oxygen because of the limited amount of KO₂. The mass of unreacted CO₂ can be calculated as: m(CO₂) = n(CO₂) x M(CO₂) = 0.656 moles x 54.09 g/mol = 35.4 g Therefore, the answer is B) 37.3 g (45.0 g - 35.4 g = 9.6 g of CO₂ was converted to O2, and 45.0 g - 9.6 g = 35.4 g of CO₂ was not converted to O2)

A Self-Contained Breathing Apparatus Contains 25 →\rightarrow→ 2K2CO3(s) + 3O2(g) A) 7

A Self-Contained Breathing Apparatus Contains 25 $\rightarrow$ 2K₂CO₃(s) + 3O₂(g)
A) 7