Question

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 self-contained 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?

Short answer

(a) 0.300 moles of O₂ are produced when 0.400 moles of KO₂ react. (b) 22.22 grams of KO₂ are needed to form 7.50 grams of O₂. (c) 6.87 grams of CO₂ are used when 7.50 grams of O₂ are produced.

Step-by-step solution

a) Moles of O₂ produced from 0.400 mol KO₂

Given the balanced chemical equation: \( 4 KO_{2} + 2 CO_{2} \rightarrow 2 K_{2} CO_{3} + 3 O_{2} \) If 0.400 mol of KO₂ reacts, we want to find the moles of O₂ produced. We will use stoichiometry to perform this calculation: \( moles\, of\, O_{2} = \frac{moles\, of\, KO_{2}}{stoichiometric\, coefficient\, of\, KO_{2}} \times stoichiometric\, coefficient\, of\, O_{2} \) \(= \frac{0.400 \,mol}{4} \times 3 \) \(= 0.300 \, mol \, O_{2} \) So, 0.300 moles of O₂ are produced when 0.400 moles of KO₂ react.

b) Grams of KO₂ needed to form 7.50 grams of O₂

First, convert 7.50 grams of O₂ to moles using its molar mass: \( moles\, of\, O_{2} = \frac{7.50 \, g}{32.00 \, g/mol} = 0.234375 \, mol \) Now, use the stoichiometry to find moles of KO₂ needed: \( moles\, of\, KO_{2} = \frac{moles\, of\, O_{2}}{stoichiometric\, coefficient\, of\, O_{2}} \times stoichiometric\, coefficient\, of\, KO_{2} \) \(= \frac{0.234375 \,mol}{3} \times 4 \) \(= 0.312500 \, mol \, KO_{2} \) Finally, convert moles of KO₂ to grams using its molar mass: \( grams\, of\, KO_{2} = 0.312500 \, mol \times 71.10 \, g/mol \) \(= 22.22 \, g \, KO_{2} \) So, 22.22 grams of KO₂ are needed to form 7.50 grams of O₂.

c) Grams of CO₂ used when 7.50 grams of O₂ are produced

We already found the moles of O₂ (0.234375 mol) for 7.50 grams of O₂ production. Now, use stoichiometry to find moles of CO₂ used: \( moles\, of\, CO_{2} = \frac{moles\, of\, O_{2}}{stoichiometric\, coefficient\, of\, O_{2}} \times stoichiometric\, coefficient\, of\, CO_{2} \) \(= \frac{0.234375 \,mol}{3} \times 2 \) \(= 0.156250 \, mol \, CO_{2} \) Finally, convert moles of CO₂ to grams using its molar mass: \( grams\, of\, CO_{2} = 0.156250 \, mol \times 44.01 \, g/mol \) \(= 6.87 \, g \, CO_{2} \) So, 6.87 grams of CO₂ are used when 7.50 grams of O₂ are produced.