The reaction between potassium superoxide, \(\mathrm{KO}_{2}\), and \(\mathrm{CO}_{2}\), $$ 4 \mathrm{KO}_{2}+2 \mathrm{CO}_{2} \longrightarrow 2 \mathrm{~K}_{2} \mathrm{CO}_{3}+3 \mathrm{O}_{2} $$ is used as a source of \(\mathrm{O}_{2}\) and absorber of \(\mathrm{CO}_{2}\) in selfcontained breathing equipment used by rescue workers. (a) How many moles of \(\mathrm{O}_{2}\) are produced when \(0.400 \mathrm{~mol}\) of \(\mathrm{KO}_{2}\) reacts in this fashion? (b) How many grams of \(\mathrm{KO}_{2}\) are needed to form \(7.50 \mathrm{~g}\) of \(\mathrm{O}_{2}\) ? (c) How many grams of \(\mathrm{CO}_{2}\) are used when \(7.50 \mathrm{~g}\) of \(\mathrm{O}_{2}\) are produced?

Step by step solution

01

(Part a: Find moles of O₂ produced)

(We are given the moles of KO₂ that react (0.400 mol) and need to find the moles of O₂ produced. To do this, we will use stoichiometry and the balanced chemical equation: \[4 \mathrm{KO}_{2}+2 \mathrm{CO}_{2} \longrightarrow 2 \mathrm{K}_{2}\mathrm{CO}_{3}+3 \mathrm{O}_{2}\] Since the balanced equation tells us that 4 moles of KO₂ produce 3 moles of O₂, we can set up a conversion factor to determine the moles of O₂ produced: moles of O₂ = moles of KO₂ × (3 moles of O₂ / 4 moles of KO₂)) Now plug in the given moles of KO₂: moles of O₂ = 0.400 mol × (3 moles of O₂ / 4 moles of KO₂))

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!