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Chapter 12: Problem 36

A self-contained breathing apparatus uses canisters containing potassium superoxide. The superoxide consumes the \(\mathrm{CO}_{2}\) exhaled by a person and replaces it with oxygen. $$ 4 \mathrm{KO}_{2}(\mathrm{s})+2 \mathrm{CO}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{K}_{2} \mathrm{CO}_{3}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{g}) $$ What mass of \(\mathrm{KO}_{2}\), in grams, is required to react with \(8.90 \mathrm{L}\) of \(\mathrm{CO}_{2}\) at \(22.0^{\circ} \mathrm{C}\) and \(767 \mathrm{mm}\) Hg?

Short Answer

Expert verified
52.20 grams of KO2 is needed.

Step by step solution

01

Convert Conditions to Standard Units

First, we need to convert the condition of carbon dioxide (CO_2) given into standard units for the calculation. The initial conditions are 22.0掳C for the temperature and 767 mmHg for the pressure.- **Convert temperature to Kelvin:**\[\text{T(K)} = \text{T(掳C)} + 273.15 = 22.0 + 273.15 = 295.15 \, \text{K}\]- **Convert pressure to atm:**\[\text{P(atm)} = \frac{767 \, \text{mmHg}}{760 \, \text{mmHg/atm}} = 1.0092 \, \text{atm}\]
02

Calculate Moles of CO2 Using Ideal Gas Law

Apply the Ideal Gas Law, \( PV = nRT \), to find the number of moles of \( \mathrm{CO}_2 \). Here, \( P = 1.0092 \, \text{atm} \), \( V = 8.90 \, \text{L} \), \( R = 0.0821 \, \text{L atm/mol K} \), and \( T = 295.15 \, \text{K} \).\[n = \frac{PV}{RT} = \frac{1.0092 \, \text{atm} \times 8.90 \, \text{L}}{0.0821 \, \text{L atm/mol K} \times 295.15 \, \text{K}} \approx 0.367 \, \text{mol of CO}_2\]This calculation gives us the moles of \( \mathrm{CO}_2 \) at the given conditions.
03

Use Stoichiometry of Reaction

Using the balanced chemical equation given:\[4 \mathrm{KO}_2 + 2 \mathrm{CO}_2 \rightarrow 2 \mathrm{K}_2\mathrm{CO}_3 + 3\mathrm{O}_2\]Note the stoichiometry, where 2 moles of \( \mathrm{CO}_2 \) react with 4 moles of \( \mathrm{KO}_2 \). Thus, dividing the moles of \( \mathrm{CO}_2 \) by 2 and multiplying by 4 gives the moles of \( \mathrm{KO}_2 \).\[\text{Moles of KO}_2 = 0.367 \, \text{mol} \times \frac{4}{2} = 0.734 \, \text{mol KO}_2\]
04

Convert Moles of KO2 to Mass

Find the molar mass of \( \mathrm{KO}_2 \) by summing atomic masses: \( \text{K} = 39.10, \quad \text{O} = 16.00 \)\[\text{Molar mass of KO}_2 = 39.10 + 2(16.00) = 71.10 \, \text{g/mol}\]Now, convert moles of \( \mathrm{KO}_2 \) to grams:\[\text{Mass of KO}_2 = 0.734 \, \text{mol KO}_2 \times 71.10 \, \text{g/mol} \approx 52.20 \, \text{g}\]Thus, the mass of \( \mathrm{KO}_2 \) required is 52.20 grams.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equations
Chemical equations are essential in chemistry as they represent how substances interact with each other. Understanding these equations is like learning a language of how chemicals talk. Each side of a chemical equation must be balanced, meaning there are equal numbers of each type of atom on both sides. This conservation of mass ensures that no atoms are lost or gained in the reaction.
In the given exercise, the chemical equation is: \[4 \mathrm{KO}_{2}(\mathrm{s})+2 \mathrm{CO}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{K}_{2} \mathrm{CO}_{3}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{g})\]
This equation tells us several things:
  • 4 moles of potassium superoxide (KO鈧�) react with 2 moles of carbon dioxide (CO鈧�).
  • It produces 2 moles of potassium carbonate (K鈧侰O鈧�) and 3 moles of oxygen gas (O鈧�).
Balancing chemical equations involves making sure each atom appears equally on both sides. This equation shows that every potassium, oxygen, and carbon atom is accounted for in the reaction.
Ideal Gas Law
The Ideal Gas Law is a cornerstone of physical chemistry, providing a relation between pressure, volume, temperature, and the number of moles of a gas. Represented by the formula \[PV = nRT\]
where:
  • \(P\) is the pressure of the gas.
  • \(V\) is the volume of the gas.
  • \(n\) is the number of moles.
  • \(R\) is the ideal gas constant.
  • \(T\) is the temperature in Kelvin.
To utilize this law, we must ensure all units are consistent, meaning temperature should be in Kelvin and pressure in atmospheres.
This was the basis for calculating the moles of CO鈧� in the exercise. By rearranging the formula to solve for \(n\), the number of moles, students can directly find how much gas they have under specific conditions.
Chemical Reactions
Chemical reactions are processes where reactants are transformed into products. Understanding reactions is key to mastering chemistry, as it explains how and why different substances interact.
For the reaction involving potassium superoxide and carbon dioxide, it's not just about watching two substances change; it's about grasping the entire process:
  • The reactants \(\mathrm{KO}_{2}\) and \(\mathrm{CO}_{2}\) combine to form new products \(\mathrm{K}_{2} \mathrm{CO}_{3}\) and \(\mathrm{O}_{2}\).
  • Every reaction has a stoichiometry, which tells us the ratio in which reactants and products appear.
In this exercise, the critical concept was recognizing the stoichiometric ratio: 4 moles of \(\mathrm{KO}_{2}\) react with 2 moles of \(\mathrm{CO}_{2}\). This ratio helped calculate the amount of potassium superoxide needed when a specific volume of carbon dioxide is given.
Understanding reactions involves not only knowing the transformations but also mastering concepts like conservation of mass, energy changes, and reaction rates.

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Most popular questions from this chapter

The gas \(\mathrm{B}_{2} \mathrm{H}_{6}\) burns in air to give \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{B}_{2} \mathrm{O}_{3}\) $$ \mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$ (a) Three gases are involved in this reaction. Place them in order of increasing molecular speed. (Assume all are at the same temperature.) (b) A \(3.26-\) I. flask contains \(B_{2} H_{6}\) at a pressure of \(256 \mathrm{mm}\) Hg and a temperature of \(25^{\circ} \mathrm{C}\). Suppose \(\mathrm{O}_{2}\) gas is added to the flask until \(\mathrm{B}_{2} \mathrm{H}_{6}\) and \(\mathrm{O}_{2}\) are in the correct stoichiometric ratio for the combustion reaction. At this point, what is the partial pressure of \(\mathrm{O}_{2} ?\)

A A study of climbers who reached the summit of Mount Everest without supplemental oxygen showed that the partial pressures of \(\mathrm{O}_{2}\) and \(\mathrm{CO}_{2}\) in their lungs were \(35 \mathrm{mm}\) Hg and \(7.5 \mathrm{mm}\) Hg, respectively. The barometric pressure at the summit was \(253 \mathrm{mm}\) Hg. Assume the lung gases are saturated with moisture at a body temperature of \(37^{\circ} \mathrm{C}\) [which means the partial pressure of water vapor in the lungs is \(P\left(\mathrm{H}_{2} \mathrm{O}\right)=47.1 \mathrm{mm} \mathrm{Hg}\) ]. If you assume the lung gases consists of only \(\mathrm{O}_{2}, \mathrm{N}_{2}, \mathrm{CO}_{2},\) and \(\mathrm{H}_{2} \mathrm{O},\) what is the partial pressure of \(\mathrm{N}_{2} ?\)

Nitrogen monoxide reacts with oxygen to give nitrogen dioxide: $$ 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g}) $$ (a) Place the three gases in order of increasing rms speed at \(298 \mathrm{K}\) (b) If you mix \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) in the correct stoichiometric ratio, and NO has a partial pressure of \(150 \mathrm{mm} \mathrm{Hg}\) what is the partial pressure of \(\mathrm{O}_{2} ?\) (c) After reaction between \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) is complete, what is the pressure of \(\mathrm{NO}_{2}\) if the NO originally had a pres. sure of \(150 \mathrm{mm} \mathrm{Hg}\) and \(\mathrm{O}_{2}\) was added in the correct stoichiometric amount?

A xenon fluoride can be prepared by heating a mixture of Xe and \(\mathrm{F}_{2}\) gases to a high temperature in a pressure-proof container. Assume that xenon gas was added to a \(0.25-\mathrm{L}\) container until its pressure reached 0.12 atm at \(0.0^{\circ} \mathrm{C}\) Fluorine gas was then added until the total pressure reached 0.72 atm at \(0.0^{\circ} \mathrm{C}\). After the reaction was complete, the xenon was consumed completely and the pressure of the \(\mathrm{F}_{2}\) remaining in the container was \(0.36 \mathrm{atm}\) at \(0.0^{\circ} \mathrm{C} .\) What is the empirical formula of the xenon fluoride?

You have a sample of helium gas at \(-33^{\circ} \mathrm{C},\) and you want to increase the average speed of helium atoms by \(10.0 \% .\) To what temperature should the gas be heated to accomplish this?
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