Problem 115 - Chemistry The Central Science

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 \(\mathrm{CO}_{2}(g) .(\mathbf{b})\) 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 equation for the reaction between $\mathrm{KO}_{2}(s)$ and $\mathrm{CO}_{2}(g)$ is: \[2 \mathrm{KO}_{2}(s) + \mathrm{CO}_{2}(g) \rightarrow \mathrm{K}_{2}\mathrm{CO}_{3}(s) + 2 \mathrm{O}_{2}(g)\] For the oxidation numbers: Oxidation: O₂²⁻ in $\mathrm{KO}_{2}$ is being oxidized, with the oxidation number increasing from -1 to 0. Reduction: C in $\mathrm{CO}_{2}$ is being reduced, with a constant oxidation number of +4. To consume 18.0 g of $\mathrm{CO}_{2}(g)$, 58.17 g of $\mathrm{KO}_{2}(s)$ is needed and 26.18 g of $\mathrm{O}_{2}(g)$ will be produced.

Step 1: Write the unbalanced equation

Initially, we write the unbalanced equation with the given reactants and products: \[ KO_2 + CO_2 \rightarrow K_2CO_3 + O_2 \]

Step 2: Balance the atoms

Now we balance the atoms in the equation: - There are 2 K atoms in the product side, so we place a coefficient of 2 before KO₂: \[ 2 KO_2 + CO_2 \rightarrow K_2CO_3 + O_2 \] - There are 4 O atoms on the reactant side, so we place a coefficient of 2 before O₂ in the product side: \[ 2 KO_2 + CO_2 \rightarrow K_2CO_3 + 2 O_2 \] Now every atom is balanced in the equation. #b) Oxidation numbers and redox elements#

Step 3: Assign oxidation numbers

We assign oxidation numbers to the atoms in the equation: In KO₂: K = +1, O₂²⁻ = -1 In CO₂: C = +4, O = -2 In K₂CO₃: K = +1, C = +4, O = -2 In O₂: O = 0

Step 4: Identify redox elements

Now we identify the elements being oxidized and reduced: Oxidation: O₂²⁻ in KO₂ loses electrons, so it is being oxidized. Its oxidation number increases from -1 to 0. Reduction: C in CO₂ gains electrons, so it is being reduced. Its oxidation number remains +4. #c) Mass of KO₂ needed and mass of O₂ produced#

Step 5: Moles of CO₂

First, we convert the given mass of CO₂ to moles. The molar mass of CO₂ is (12.01 + 2 × 16.00) g/mol: \[ \text{moles of CO₂} = \frac{18.0\,\text{g}}{44.01\,\text{g/mol}} = 0.409\,\text{mol} \]

Step 6: Moles of KO₂ needed

From the balanced equation, we know that 2 moles of KO₂ react with 1 mole of CO₂. So, we need: \[ \text{moles of KO₂} = 2 \times 0.409\,\text{mol} = 0.818\,\text{mol} \]

Step 7: Mass of KO₂ needed

Now we convert the moles of KO₂ to mass. The molar mass of KO₂ is (39.10 + 2 × 16.00) g/mol: \[ \text{mass of KO₂} = 0.818\,\text{mol} \times 71.10\,\text{g/mol} = 58.17\,\text{g} \]

Step 8: Moles of O₂ produced

From the balanced equation, 2 moles of KO₂ produce 2 moles of O₂. So, the moles of O₂ produced are: \[ \text{moles of O₂} = 0.818\,\text{mol} \]

Step 9: Mass of O₂ produced

Finally, we convert the moles of O₂ to mass. The molar mass of O₂ is 2 × 16.00 g/mol: \[ \text{mass of O₂} = 0.818\,\text{mol} \times 32.00\,\text{g/mol} = 26.18\,\text{g} \] In conclusion, 58.17 g of KO₂ are needed to consume 18.0 g of CO₂, and 26.18 g of O₂ will be produced.

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry that ensures the law of conservation of mass is followed. This law states that matter cannot be created or destroyed in a chemical reaction. Therefore, the number of atoms for each element must be the same on both sides of the reaction.

To balance an equation, start by writing the unbalanced equation showing the reactants and products. Identify the atoms involved and compare the number of atoms for each element on both sides. Use coefficients, which are numbers placed in front of compounds, to ensure that the number of atoms of each element is equal on both sides. Never change the subscripts of compounds, as this changes the identity of the substances involved.

For example, in the reaction with potassium superoxide and carbon dioxide, the initial unbalanced equation is:
\[ KO_2 + CO_2 \rightarrow K_2CO_3 + O_2 \]
Balancing starts with matching potassium and oxygen atoms, adjusting coefficients accordingly:
\[ 2 KO_2 + CO_2 \rightarrow K_2CO_3 + 2 O_2 \]
This ensures that there are equal numbers of K and O atoms on each side, obeying the law of conservation of mass and completing the balancing process.

Oxidation-Reduction (Redox) Reactions

Oxidation-reduction, or redox, reactions involve the transfer of electrons between species. In such reactions, oxidation refers to the loss of electrons, and reduction refers to the gain of electrons. The species that donates electrons is oxidized, and the species that accepts electrons is reduced.

To identify redox reactions, one must assign oxidation numbers to each atom in the reactants and products. Oxidation numbers represent the potential charge an atom would have in a molecule, assuming that electrons are transferred completely.

For example, in the reaction between KO₂ and CO₂, the oxidation numbers help identify the changes:

• In KO₂: K = +1, each O = -1/2 (since O₂²⁻ = -1).
• In CO₂: C = +4, each O = -2.
• In K₂CO₃: K = +1, C = +4, each O = -2.
• In O₂: O = 0.

In this reaction, the O₂²⁻ ion is oxidized as it loses electrons and forms O₂ gas. The oxidation number of oxygen changes from -1 to 0. Carbon remains at an oxidation number of +4, so it is neither oxidized nor reduced. This step of determining the redox elements is crucial for understanding electron flow in chemical reactions.

Stoichiometry

Stoichiometry is the quantitative aspect of chemical equations. It involves calculations based on the balanced equation to determine the amounts of reactants and products involved in a chemical reaction. Using stoichiometry, one can predict how much reactant is needed to produce a certain amount of product or the quantity of product formed from a given amount of reactant.

A stoichiometric calculation typically begins with the moles of one substance. A balanced chemical equation then shows the mole ratios between reactants and products, which allows the conversion to other substances in the reaction. Finally, with the use of molar masses, these mole quantities can be converted into masses.

For instance, to calculate the mass of KO₂ needed to consume 18.0 grams of CO₂ (given that the molar mass of CO₂ is 44.01 g/mol) we first find the moles of CO₂:
\[ \text{moles of CO₂} = \frac{18.0\text{g}}{44.01\text{g/mol}} = 0.409 \text{mol} \]
Using the balanced chemical equation, derive the moles of KO₂ needed and convert to mass. Likewise, calculate the mass of O₂ produced. This kind of stoichiometric analysis is essential for practical applications like the design of oxygen masks for firefighters, as in the given exercise.