Question

Potassium superoxide, \(\mathrm{KO}_{2}\), is often used in oxygen masks (such as those used by firefighters) because \(\mathrm{KO}_{2}\) reacts with \(\mathrm{CO}_{2}\) to release molecular oxygen. Experiments indicate that \(2 \mathrm{~mol}\) of \(\mathrm{KO}_{2}(s)\) react with each mole of \(\mathrm{CO}_{2}(g)\). (a) The products of the reaction are \(\mathrm{K}_{2} \mathrm{CO}_{3}(s)\) and \(\mathrm{O}_{2}(g)\). Write a balanced equation for the reaction between \(\mathrm{KO}_{2}(s)\) and \(\mathrm{CO}_{2}(g)\). (b) Indicate the oxidation number for each atom involved in the reaction in part (a). What elements are being oxidized and reduced? (c) What mass of \(\mathrm{KO}_{2}(s)\) is needed to consume \(18.0 \mathrm{~g} \mathrm{CO}_{2}(\mathrm{~g})\) ? What mass of \(\mathrm{O}_{2}(g)\) is produced during this reaction?

Short answer

The balanced chemical equation for the reaction between potassium superoxide (\(\mathrm{KO}_{2}\)) and carbon dioxide (\(\mathrm{CO}_{2}\)) is: \(2\,\mathrm{KO}_{2} + \mathrm{CO}_{2} \rightarrow \mathrm{K}_{2}\mathrm{CO}_{3} + \dfrac{3}{2}\,\mathrm{O}_{2}\). The oxidation number of oxygen in \(\mathrm{KO}_{2}\) increases from \(-1/2\) to \(-2\), so oxygen is being reduced. A total of \(58.2\,\mathrm{g}\) of \(\mathrm{KO}_{2}\) is needed to consume \(18.0\,\mathrm{g}\) \(\mathrm{CO}_{2}\), and during this reaction, \(19.6\,\mathrm{g}\) of \(\mathrm{O}_{2}\) is produced.

Step-by-step solution

Write the unbalanced equation

The given unbalanced chemical equation involves potassium superoxide(\(\mathrm{KO}_{2}\)), carbon dioxide(\(\mathrm{CO}_{2}\)), potassium carbonate(\(\mathrm{K}_{2}\mathrm{CO}_{3}\)), and molecular oxygen(\(\mathrm{O}_{2}\)) as reactants and products. So the equation is: \[\mathrm{KO}_{2} + \mathrm{CO}_{2} \rightarrow \mathrm{K}_{2}\mathrm{CO}_{3} + \mathrm{O}_{2}\]

Balance the chemical equation

To balance the equation, we need to have an equal number of each type of atom on both sides of the equation. It is given that \(2 \mathrm{~mol}\) of \(\mathrm{KO}_{2}\) react with each mole of \(\mathrm{CO}_{2}\). Therefore, the balanced chemical equation is: \[2\,\mathrm{KO}_{2} + \mathrm{CO}_{2} \rightarrow \mathrm{K}_{2}\mathrm{CO}_{3} + \dfrac{3}{2}\,\mathrm{O}_{2}\] This balanced equation indicates that for every \(2\) moles of \(\mathrm{KO}_{2}\), \(1\) mole of \(\mathrm{CO}_{2}\) reacts to form \(1\) mole of \(\mathrm{K}_{2}\mathrm{CO}_{3}\) and \(\dfrac{3}{2}\) moles of \(\mathrm{O}_{2}\).

Determine oxidation numbers and identify oxidized and reduced elements

We need to find the oxidation numbers of each atom in the compounds and determine which elements are being oxidized and reduced. Let's find their oxidation numbers: Potassium(K): \(+1\) Oxygen(O) in \(\mathrm{KO}_{2}\): \(-1/2\) Oxygen(O) in \(\mathrm{CO}_{2}\) and \(\mathrm{K}_{2}\mathrm{CO}_{3}\): \(-2\) Carbon(C) in \(\mathrm{CO}_{2}\) and \(\mathrm{K}_{2}\mathrm{CO}_{3}\): \(+4\) As the oxidation number of oxygen in \(\mathrm{KO}_{2}\) increases from \(-1/2\) to \(-2\), oxygen is being reduced. On the other hand, the oxidation number of potassium remains constant at \(+1\). Thus, the elements being oxidized and reduced in the reaction are oxygen and potassium.

Calculate mass of \(\mathrm{KO}_{2}\) needed to consume \(18.0\,\mathrm{g}\) \(\mathrm{CO}_{2}\)

To calculate the mass of \(\mathrm{KO}_{2}\) needed, first determine the number of moles of \(\mathrm{CO}_{2}\) in the reaction. Use the molar mass of \(\mathrm{CO}_{2}\): \(1\,\mathrm{mol}\) \(\mathrm{CO}_{2} = 44.01\,\mathrm{g}\) Number of moles of \(\mathrm{CO}_{2}\): \[\dfrac{18.0\,\mathrm{g}}{44.01\,\mathrm{g/mol}} = 0.409\,\mathrm{mol}\, \mathrm{CO}_{2}\] Now, use the balanced equation to calculate the moles of \(\mathrm{KO}_{2}\) required: \[2\,\mathrm{mol}\,\mathrm{KO}_{2} = 0.409\,\mathrm{mol}\,\mathrm{CO}_{2}\] Moles of \(\mathrm{KO}_{2}\) needed: \[0.409\,\mathrm{mol}\,\mathrm{CO}_{2} \times \dfrac{2\,\mathrm{mol}\,\mathrm{KO}_{2}}{1\,\mathrm{mol}\,\mathrm{CO}_{2}} = 0.818\,\mathrm{mol}\,\mathrm{KO}_{2}\] Finally, calculate the mass of \(\mathrm{KO}_{2}\) required (using the molar mass of \(\mathrm{KO}_{2}\): \(1\,\mathrm{mol}\) \(\mathrm{KO}_{2} = 71.1\,\mathrm{g}\)): Mass of \(\mathrm{KO}_{2}\) needed: \[0.818\,\mathrm{mol}\,\mathrm{KO}_{2} \times \dfrac{71.1\,\mathrm{g}}{1\,\mathrm{mol}\,\mathrm{KO}_{2}} = 58.2\,\mathrm{g}\,\mathrm{KO}_{2}\]

Calculate mass of \(\mathrm{O}_{2}\) produced during the reaction

To calculate the mass of \(\mathrm{O}_{2}\) produced, first determine the number of moles of \(\mathrm{O}_{2}\) produced using the balanced equation: \[0.409\,\mathrm{mol}\,\mathrm{CO}_{2} \times \dfrac{3/2\,\mathrm{mol}\,\mathrm{O}_{2}}{1\,\mathrm{mol}\,\mathrm{CO}_{2}} = 0.614\,\mathrm{mol}\,\mathrm{O}_{2}\] Now, calculate the mass of \(\mathrm{O}_{2}\) produced using the molar mass of \(\mathrm{O}_{2}\): \(1\,\mathrm{mol}\) \(\mathrm{O}_{2} = 32.0\,\mathrm{g}\) Mass of \(\mathrm{O}_{2}\) produced: \[0.614\,\mathrm{mol}\,\mathrm{O}_{2} \times \dfrac{32.0\,\mathrm{g}}{1\,\mathrm{mol}\,\mathrm{O}_{2}} = 19.6\,\mathrm{g}\,\mathrm{O}_{2}\] In conclusion, \(58.2\,\mathrm{g}\) of \(\mathrm{KO}_{2}\) is needed to consume \(18.0\,\mathrm{g}\) \(\mathrm{CO}_{2}\), and during this reaction, \(19.6\,\mathrm{g}\) of \(\mathrm{O}_{2}\) is produced.

Key concepts

Balanced Equations

In chemical reactions, balanced equations are essential as they show the precise ratio in which reactants combine to form products. Balancing a chemical equation means ensuring that the same number of each type of atom appears on both sides of the equation. This is crucial due to the Law of Conservation of Mass, which states that mass is neither created nor destroyed in a chemical reaction.

To balance an equation, you adjust the coefficients (the numbers in front of the formulas) so that each element has the same total number on both sides. For the reaction between potassium superoxide (\(\mathrm{KO}_{2}\)) and carbon dioxide (\(\mathrm{CO}_{2}\)), we initially write the equation as:

• \(\mathrm{KO}_{2} + \mathrm{CO}_{2} \rightarrow \mathrm{K}_{2}\mathrm{CO}_{3} + \mathrm{O}_{2}\)

Then, we balance the equation by finding that 2 moles of \(\mathrm{KO}_{2}\) react with 1 mole of \(\mathrm{CO}_{2}\) to yield 1 mole of \(\mathrm{K}_{2}\mathrm{CO}_{3}\) and \(\dfrac{3}{2}\) moles of \(\mathrm{O}_{2}\). Adjusting these coefficients balances the equation and represents the stoichiometry of the reaction correctly.

Oxidation States

Oxidation states help us understand how electrons are transferred between elements in a chemical reaction. The oxidation state (or number) is the hypothetical charge an atom would have if all bonds were ionic. In the context of the exercise, it's vital to determine these states to identify oxidation-reduction processes (redox reactions).

For \(\mathrm{K}_{2}\mathrm{CO}_{3}\) formation, we consider oxidation numbers:

• Potassium (K) in \(\mathrm{KO}_{2}\) and \(\mathrm{K}_{2}\mathrm{CO}_{3}\): \(+1\)
• Oxygen (O) in \(\mathrm{KO}_{2}\): \(-1/2\) and in other compounds \(-2\)
• Carbon (C) in \(\mathrm{CO}_{2}\) and \(\mathrm{K}_{2}\mathrm{CO}_{3}\): \(+4\)

Oxygen goes from an oxidation state of \(-1/2\) to \(-2\), indicating a gain of electrons—reduction. Potassium's oxidation state remains constant, implying it is not involved in the oxidation-reduction process. This analysis helps track shifts in electron density and identify elements that are oxidized or reduced in a reaction.

Stoichiometry

Stoichiometry is a fundamental concept that involves the calculation of reactants and products in chemical reactions. It allows chemists to predict how much of each substance is needed or produced under certain conditions. The key is to use the mole concept and balanced equations effectively.

To solve the exercise, we start with the balanced equation:

• \( 2\,\mathrm{KO}_{2} + \mathrm{CO}_{2} \rightarrow \mathrm{K}_{2}\mathrm{CO}_{3} + \dfrac{3}{2}\,\mathrm{O}_{2} \)

Given 18.0 grams of \(\mathrm{CO}_{2}\), we convert this mass to moles using the molar mass (44.01 g/mol). We find it equals 0.409 moles of \(\mathrm{CO}_{2}\). From the balanced equation, we know that 2 moles of \(\mathrm{KO}_{2}\) are required per mole of \(\mathrm{CO}_{2}\).

Multiplying gives 0.818 moles of \(\mathrm{KO}_{2}\) necessary. Finally, using the molar mass of \(\mathrm{KO}_{2}\) (71.1 g/mol), we determine that 58.2 grams of \(\mathrm{KO}_{2}\) are required. Calculating the produced \(\mathrm{O}_{2}\), from the stoichiometry again, yields 19.6 grams of \(\mathrm{O}_{2}\). This systematic approach ensures accuracy in predicting the outcomes of chemical reactions.