91Ó°ÊÓ

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?

Step by step solution

01

a) Moles of Oâ‚‚ produced from 0.400 mol KOâ‚‚

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 KOâ‚‚ reacts, we want to find the moles of Oâ‚‚ 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 Oâ‚‚ are produced when 0.400 moles of KOâ‚‚ react.

02

b) Grams of KOâ‚‚ needed to form 7.50 grams of Oâ‚‚

First, convert 7.50 grams of Oâ‚‚ 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 KOâ‚‚ 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 KOâ‚‚ 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 KOâ‚‚ are needed to form 7.50 grams of Oâ‚‚.

03

c) Grams of COâ‚‚ used when 7.50 grams of Oâ‚‚ are produced

We already found the moles of Oâ‚‚ (0.234375 mol) for 7.50 grams of Oâ‚‚ production. Now, use stoichiometry to find moles of COâ‚‚ 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 COâ‚‚ 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 COâ‚‚ are used when 7.50 grams of Oâ‚‚ are produced.

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