/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 156 A steel cylinder contains \(5.00... [FREE SOLUTION] | 91Ó°ÊÓ

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A steel cylinder contains \(5.00\) moles of graphite (pure carbon) and \(5.00\) moles of \(\mathrm{O}_{2}\). The mixture is ignited and all the graphite reacts. Combustion produces a mixture of \(\mathrm{CO}\) gas and \(\mathrm{CO}_{2}\) gas. After the cylinder has cooled to its original temperature, it is found that the pressure of the cylinder has increased by \(17.0 \% .\) Calculate the mole fractions of \(\mathrm{CO}, \mathrm{CO}_{2},\) and \(\mathrm{O}_{2}\) in the final gaseous mixture.

Short Answer

Expert verified
The mole fractions of CO, COâ‚‚, and Oâ‚‚ in the final gaseous mixture are approximately 0.776, 0.159, and 0.079, respectively.

Step by step solution

01

Write balanced chemical equations for the combustion reactions

There are two possible chemical reactions that can occur here: the reaction of carbon with oxygen to form carbon monoxide (CO) and the reaction of carbon with oxygen to form carbon dioxide (CO₂). Let's write both balanced chemical equations: 1. C + 1/2 O₂ → CO 2. C + O₂ → CO₂
02

Calculate the initial amount of Oâ‚‚ in the mixture

Since we start with 5 moles of graphite (pure carbon) and 5 moles of Oâ‚‚, we have a total of 5 moles of oxygen gas available for the formation of CO and COâ‚‚.
03

Calculate the maximum pressure increase of the cylinder

In a scenario where all 5 moles of oxygen reacts with all 5 moles of carbon to form CO, we will have the maximum increase in the number of moles of the mixture because it consumes half as much oxygen as the formation of COâ‚‚. Therefore, the maximum pressure increase can be calculated as: Maximum Pressure Increase = (Final moles - Initial moles) / Initial moles = (5 + 5 - 5) / 5 = 1
04

Calculate the actual pressure increase

From the given data, the pressure of the cylinder has increased by 17%. We can calculate the actual pressure increase as follows: Actual Pressure Increase = (Final moles - Initial moles) / Initial moles = 0.17
05

Determine moles of CO and COâ‚‚ formed

Let x be the moles of COâ‚‚ formed during the combustion. Then, the moles of CO formed will be 5 - x. We can use the information from Step 4 to write the equation: (5 - x + x - 5) / 5 = 0.17 Solve for x: x = 0.17 * 5 = 0.85
06

Calculate the moles of each gas in the final mixture

Now that we have found the moles of COâ‚‚ (x = 0.85), we can calculate the moles of CO and Oâ‚‚ as well: Moles of CO = 5 - x = 5 - 0.85 = 4.15 Moles of COâ‚‚ = x = 0.85 Moles of Oâ‚‚ = Initial moles of Oâ‚‚ - moles of Oâ‚‚ used in the reaction = 5 - (0.85 * 1 + 4.15 * 0.5) = 0.425
07

Calculate the mole fractions for each component in the mixture

With the moles of each gas in the final mixture, we can now find their mole fractions: Mole Fraction of CO = Moles of CO / Total moles = 4.15 / (4.15 + 0.85 + 0.425) = 0.7756 Mole Fraction of COâ‚‚ = Moles of COâ‚‚ / Total moles = 0.85 / (4.15 + 0.85 + 0.425) = 0.1587 Mole Fraction of Oâ‚‚ = Moles of Oâ‚‚ / Total moles = 0.425 / (4.15 + 0.85 + 0.425) = 0.0793 So, the mole fractions of CO, COâ‚‚, and Oâ‚‚ in the final gaseous mixture are approximately 0.776, 0.159, and 0.079, respectively.

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Key Concepts

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

Combustion Reactions
Combustion reactions are chemical processes where a fuel reacts with an oxidizer, usually oxygen, to release energy in the form of heat and light. Typically involving hydrocarbons or carbon-based substances like graphite (pure carbon), these exothermic reactions are fundamental in various applications, from engines to heating systems.

Let's consider our textbook example, where graphite undergoes combustion in the presence of oxygen. Depending on the conditions, graphite can form either carbon monoxide (CO) or carbon dioxide (COâ‚‚). Both reactions can be expressed through chemical equations:
  • C + 1/2 Oâ‚‚ → CO
  • C + Oâ‚‚ → COâ‚‚
These balanced equations reflect the law of conservation of mass, where the number of atoms of each element is the same before and after the reaction. They are pivotal in solving for the products formed in a combustion process.
Chemical Equations
Chemical equations are representations of a chemical reaction using symbols and formulas. For a balanced chemical equation, the number of atoms for each element must be equal on both sides of the equation. Achieving this balance is crucial in stoichiometry, which is the calculation of reactants and products in chemical reactions.

In the case of our steel cylinder example, we must consider two equations for the combustion of carbon. The balanced chemical equations guide us on how to calculate the moles of each reactant and product. By understanding the stoichiometry, we are able to derive that for every mole of carbon burned, either one mole of CO is produced or one mole of COâ‚‚, depending on the reaction path. With this understanding, we can predict the composition and quantities of the compounds in the final gaseous state.
Gas Laws
The gas laws are a set of laws that describe the relationship between the pressure, volume, temperature, and amount of gas. For educational purposes, let's focus on how pressure relates to the amount of gas. Boyle's Law, Charles's Law, and Avogadro's Law are parts of the ideal gas law equation, expressed as PV = nRT, where P is pressure, V is volume, n is the amount of gas in moles, R is the ideal gas constant, and T is temperature in Kelvin.

When examining the increase in pressure after the combustion reaction in our steel cylinder, the temperature and volume are constant, meaning any pressure change must be due to a change in the number of moles of gas. For a fixed amount of gas at constant temperature, higher pressure suggests an increased number of moles post-reaction, leading us to calculate the specific pressures based on the stoichiometric outcomes of the combustion process. By applying gas laws to the change in pressure data, we can deduce the composition of the gas mixture after the reaction.

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Most popular questions from this chapter

The oxides of Group \(2 \mathrm{A}\) metals (symbolized by M here) react with carbon dioxide according to the following reaction: $$\mathrm{MO}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{MCO}_{3}(s)$$ A \(2.85\) \(-\mathrm{g}\) sample containing only \(\mathrm{MgO}\) and \(\mathrm{CuO}\) is placed in a 3.00 -L container. The container is filled with \(\mathrm{CO}_{2}\) to a pressure of \(740 .\) torr at \(20 .^{\circ} \mathrm{C}\). After the reaction has gone to completion, the pressure inside the flask is \(390 .\) torr at \(20 .^{\circ} \mathrm{C}\). What is the mass percent of \(MgO\) in the mixture? Assume that only the \(MgO\) reacts with \(\mathrm{CO}_{2}\)

A steel cylinder contains \(150.0\) moles of argon gas at a temperature of \(25^{\circ} \mathrm{C}\) and a pressure of \(8.93 \mathrm{MPa}\). After some argon has been used, the pressure is \(2.00 \mathrm{MPa}\) at a temperature of \(19^{\circ} \mathrm{C}\). What mass of argon remains in the cylinder?

Consider separate \(1.0-\mathrm{L}\) gaseous samples of \(\mathrm{He}, \mathrm{N}_{2},\) and \(\mathrm{O}_{2}\) all at \(\mathrm{STP}\) and all acting ideally. Rank the gases in order of increasing average kinetic energy and in order of increasing average velocity.

An ideal gas is contained in a cylinder with a volume of \(5.0 \times 10^{2} \mathrm{mL}\) at a temperature of \(30 .^{\circ} \mathrm{C}\) and a pressure of \(710.\) torr. The gas is then compressed to a volume of \(25 \mathrm{mL}\) and the temperature is raised to \(820 .^{\circ} \mathrm{C}\). What is the new pressure of the gas?

A tank contains a mixture of \(52.5 \mathrm{g}\) oxygen gas and \(65.1 \mathrm{g}\) carbon dioxide gas at \(27^{\circ} \mathrm{C}\). The total pressure in the tank is \(9.21\) atm. Calculate the partial pressures of each gas in the container.

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