Problem 115 - Chemistry: The Central Science in SI Units, Global Edition

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 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 \(\mathrm{CO}_{2}(g) .(\mathbf{b})\) 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 \(18.0 \mathrm{~g} \mathrm{CO}_{2}(g)\) ? What mass of \(\mathrm{O}_{2}(g)\) is produced during this reaction?

Step-by-Step Solution

Answer

The balanced chemical equation for the reaction between \(\mathrm{KO}_{2}\) and \(\mathrm{CO}_{2}\) is: \[2 \mathrm{KO}_{2}(s) + \mathrm{CO}_{2}(g) \rightarrow \mathrm{K}_{2}\mathrm{CO}_{3}(s) + \mathrm{O}_{2}(g)\] In the reaction, oxygen is being reduced as its oxidation number changes from -1/2 in \(\mathrm{KO}_{2}\) to 0 in \(\mathrm{O}_{2}\). To consume 18.0 g of \(\mathrm{CO}_{2}\), 58.18 g of \(\mathrm{KO}_{2}\) is required, and during the reaction, 13.09 g of \(\mathrm{O}_{2}\) is produced.

Step 1: Write the balanced chemical equation

We are given that 2 moles of \(\mathrm{KO}_{2}\) react with 1 mole of \(\mathrm{CO}_{2}\), and the products are \(\mathrm{K}_{2}\mathrm{CO}_{3}\) and \(\mathrm{O}_{2}\). Based on these initial reactants and products, the equation would look like this: \[2 \mathrm{KO}_{2}(s) + \mathrm{CO}_{2}(g) \rightarrow \mathrm{K}_{2}\mathrm{CO}_{3}(s) + \mathrm{O}_{2}(g)\] This equation is already balanced, as there are the same number of each type of atom on both sides of the equation.

Step 2: Determine the oxidation numbers

In order to identify the elements that are being oxidized and reduced, we first need to determine the oxidation numbers of each element. For \(\mathrm{KO}_{2}\): K: +1 (potassium is an alkali metal and always has an oxidation number of +1) O: -1/2 (oxygen in superoxides has an oxidation number of -1/2) For \(\mathrm{CO}_{2}\): C: +4 (carbon in \(\mathrm{CO}_{2}\) has an oxidation number of +4) O: -2 (oxygen in most compounds has an oxidation number of -2) For \(\mathrm{K}_{2}\mathrm{CO}_{3}\): K: +1 C: +4 O: -2 For \(\mathrm{O}_{2}\): O: 0 (in its elemental form, oxygen has an oxidation number of 0)

Step 3: Identify the elements being oxidized and reduced

Using the oxidation numbers we found in Step 2, we can now identify the elements being oxidized and reduced. Potassium's oxidation number remains the same, so it is not being oxidized or reduced. Oxygen's oxidation number changes from -1/2 in \(\mathrm{KO}_{2}\) to 0 in \(\mathrm{O}_{2}\), so it is being reduced. Carbon's oxidation number remains the same, so it is not being oxidized or reduced.

Step 4: Calculate the mass of \(\mathrm{KO}_{2}\) needed

We are given that 18.0 g of \(\mathrm{CO}_{2}\) needs to react with \(\mathrm{KO}_{2}\). We can use stoichiometry to find the mass of \(\mathrm{KO}_{2}\) needed. First, convert the mass of \(\mathrm{CO}_{2}\) to moles: \[18.0 \, \text{g} \, \mathrm{CO}_{2} \cdot \frac{1 \, \text{mol} \, \mathrm{CO}_{2}}{44.01 \, \text{g} \, \mathrm{CO}_{2}} = 0.409 \, \text{mol}\, \mathrm{CO}_{2} \] Now, using the balanced equation from Step 1, we can find the moles of \(\mathrm{KO}_{2}\) needed: \[0.409 \, \text{mol}\, \mathrm{CO}_{2} \cdot \frac{2 \, \text{mol} \, \mathrm{KO}_{2}}{1 \, \text{mol} \, \mathrm{CO}_{2}} = 0.818 \, \text{mol}\, \mathrm{KO}_{2} \] Finally, convert the moles of \(\mathrm{KO}_{2}\) to grams: \[0.818 \, \text{mol}\, \mathrm{KO}_{2} \cdot \frac{71.10 \, \text{g} \, \mathrm{KO}_{2}}{1 \, \text{mol} \, \mathrm{KO}_{2}} = 58.18\text{g}\, \mathrm{KO}_{2} \] Therefore, 58.18 g of \(\mathrm{KO}_{2}\) is needed to react with 18.0 g of \(\mathrm{CO}_{2}\).

Step 5: Calculate the mass of \(\mathrm{O}_{2}\) produced

Using stoichiometry and the balanced equation, we can now find the mass of \(\mathrm{O}_{2}\) produced. Convert the moles of \(\mathrm{CO}_{2}\) that reacted to moles of produced \(\mathrm{O}_{2}\): \[0.409 \, \text{mol}\, \mathrm{CO}_{2} \cdot \frac{1 \, \text{mol} \, \mathrm{O}_{2}}{1 \, \text{mol} \, \mathrm{CO}_{2}} = 0.409 \, \text{mol}\, \mathrm{O}_{2} \] Finally, convert the moles of \(\mathrm{O}_{2}\) to grams: \[0.409 \, \text{mol}\, \mathrm{O}_{2} \cdot \frac{32.00 \, \text{g} \, \mathrm{O}_{2}}{1 \, \text{mol} \, \mathrm{O}_{2}} = 13.09\text{g}\, \mathrm{O}_{2} \] Therefore, 13.09 g of \(\mathrm{O}_{2}\) is produced during the reaction.

Key Concepts

Balancing Chemical Equations

In chemistry, balancing chemical equations is a crucial skill that ensures both sides of a reaction reflect the conservation of mass, meaning the atoms are neither created nor destroyed in a reaction. In the case of the given reaction between potassium superoxide \(\mathrm{KO}_2(s)\) and carbon dioxide \(\mathrm{CO}_2(g)\), the balanced equation is: \[2\ \mathrm{KO}_2(s) + \mathrm{CO}_2(g) \rightarrow \mathrm{K}_2\mathrm{CO}_3(s) + \mathrm{O}_2(g)\] This equation confirms that the number of each type of atom on the left (reactants) matches the number on the right (products).

• There are 2 potassium atoms, 3 oxygen atoms, and 1 carbon atom on both sides.
• This balancing ensures a proper stoichiometric ratio, serving as the foundation for further calculations like determining reactants or products needed in a reaction.

Understanding how to systematically approach balancing involves counting each atom in the reactants and products, adjusting coefficients, and verifying their equivalence on both sides of the equation. Be sure to adjust only the coefficients (the numbers before substances) and not the subscripts (the small numbers within chemical formulas) to maintain the correct makeup of each compound.

Stoichiometry

Stoichiometry is the branch of chemistry that pertains to measuring and calculating the relative quantities of reactants and products in chemical reactions. In the demonstrated reaction, stoichiometry helps determine how much \(\mathrm{KO}_2\) is needed to react with a known mass of \(\mathrm{CO}_2\). Given 18.0 g of \(\mathrm{CO}_2\), we first convert this mass to moles using its molar mass:
\[18.0 \, \text{g} \, \mathrm{CO}_2 \cdot \frac{1 \, \text{mol} \, \mathrm{CO}_2}{44.01 \, \text{g} \, \mathrm{CO}_2} = 0.409 \, \text{mol}\, \mathrm{CO}_2\] Using the balanced equation, we find the stoichiometric ratio between \(\mathrm{CO}_2\) and \(\mathrm{KO}_2\):

• For each 1 mole of \(\mathrm{CO}_2\), 2 moles of \(\mathrm{KO}_2\) are needed.

Thus, the moles of \(\mathrm{KO}_2\) are:
\[0.409 \, \text{mol}\, \mathrm{CO}_2 \cdot \frac{2 \, \text{mol} \, \mathrm{KO}_2}{1 \, \text{mol} \, \mathrm{CO}_2} = 0.818 \, \text{mol}\, \mathrm{KO}_2\] Converting this to grams using the molar mass of \(\mathrm{KO}_2\) gives us the required mass of \(\mathrm{KO}_2\):
\[0.818 \, \text{mol}\, \mathrm{KO}_2 \cdot \frac{71.10 \, \text{g} \, \mathrm{KO}_2}{1 \, \text{mol} \, \mathrm{KO}_2} = 58.18\text{g}\, \mathrm{KO}_2\]The mass of oxygen produced can also be determined using similar calculations by tracking moles through the balanced reaction and converting to grams.

Oxidation Numbers

Oxidation numbers are a way to keep track of electrons in chemical reactions, especially in redox reactions. For the reaction involving \(\mathrm{KO}_2\) and \(\mathrm{CO}_2\), understanding oxidation numbers is key to identifying elements that undergo oxidation or reduction.
- In \(\mathrm{KO}_2\), the oxidation number for potassium (K) is +1 since it is an alkali metal. For oxygen, the unusual -1/2 charge is present as it forms a superoxide. - In \(\mathrm{CO}_2\), carbon holds an oxidation state of +4, whereas oxygen is typically -2. - When \(\mathrm{K}_2\mathrm{CO}_3\) is formed, both potassium and carbon maintain their oxidation states: potassium remains +1 and carbon remains +4. - In \(\mathrm{O}_2\), the oxygen is in its elemental state at 0. The shift from -1/2 in \(\mathrm{KO}_2\) to 0 in \(\mathrm{O}_2\) indicates reduction (gain of electrons). Similarly, oxidation numbers can show whether specific elements in a reaction undergo oxidation (loss of electrons), crucial for understanding redox mechanisms.