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Chapter 4: Problem 80

Reaction of pentane, \(\mathrm{C}_{5} \mathrm{H}_{12},\) with oxygen, \(\mathrm{O}_{2},\) gives carbon dioxide and water. (a) Write a balanced equation for this reaction. (b) In this reaction, what is oxidized and what is reduced? (c) What is the oxidizing agent, and what is the reducing agent?

Short Answer

Expert verified
(a) \(\mathrm{C}_{5}\mathrm{H}_{12} + 8\mathrm{O}_{2} \rightarrow 5\mathrm{CO}_{2} + 6\mathrm{H}_{2}\mathrm{O}\); (b) \(\mathrm{C}_{5}\mathrm{H}_{12}\) is oxidized, \(\mathrm{O}_2\) is reduced; (c) \(\mathrm{O}_2\) is the oxidizing agent, \(\mathrm{C}_{5}\mathrm{H}_{12}\) is the reducing agent.

Step by step solution

01

Write the Unbalanced Equation

To start, recognize the type of reaction. Here, pentane (\(\mathrm{C}_5\mathrm{H}_{12}\)) combusts in the presence of oxygen (\(\mathrm{O}_2\)) to produce carbon dioxide (\(\mathrm{CO}_2\)) and water (\(\mathrm{H}_2\mathrm{O}\)). The unbalanced equation is:\[\mathrm{C}_{5}\mathrm{H}_{12} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}\]
02

Balance Carbon Atoms

Ensure that the number of carbon atoms on each side of the equation is equal. In this case, there are 5 carbon atoms in pentane, so we need 5 molecules of carbon dioxide:\[\mathrm{C}_{5}\mathrm{H}_{12} + \mathrm{O}_{2} \rightarrow 5\mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}\]
03

Balance Hydrogen Atoms

Next, balance the hydrogen atoms. Pentane has 12 hydrogen atoms, so we need 6 water molecules to balance the hydrogen:\[\mathrm{C}_{5}\mathrm{H}_{12} + \mathrm{O}_{2} \rightarrow 5\mathrm{CO}_{2} + 6\mathrm{H}_{2}\mathrm{O}\]
04

Balance Oxygen Atoms

Balance the oxygen atoms last. The products have a total of 16 oxygen atoms (10 from CO2 and 6 from H2O), so 8 molecules of O2 are required:\[\mathrm{C}_{5}\mathrm{H}_{12} + 8\mathrm{O}_{2} \rightarrow 5\mathrm{CO}_{2} + 6\mathrm{H}_{2}\mathrm{O}\]
05

Identify Oxidized and Reduced Species

In combustion reactions, hydrocarbons are oxidized. Pentane (\(\mathrm{C}_5\mathrm{H}_{12}\)) is oxidized to carbon dioxide (\(\mathrm{CO}_2\)). Oxygen (\(\mathrm{O}_2\)) is reduced to water (\(\mathrm{H}_2\mathrm{O}\)).
06

Determine Oxidizing and Reducing Agents

The substance that is reduced (oxygen) is the oxidizing agent, while the substance that is oxidized (pentane) is the reducing agent.

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Key Concepts

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

Oxidation-Reduction
In chemistry, oxidation-reduction reactions, commonly called redox reactions, involve the transfer of electrons between two species. An oxidation occurs when a molecule, atom, or ion loses electrons, whereas a reduction occurs when a molecule, atom, or ion gains electrons. In combustion reactions, such as the one between pentane (\(\mathrm{C}_5\mathrm{H}_{12}\)) and oxygen (\(\mathrm{O}_2\)), these processes happen simultaneously.
For the reaction in question, pentane is oxidized, meaning it loses electrons, and oxygen is reduced, meaning it gains electrons. Such reactions are crucial in many industrial and biological processes, as they are a major source of energy.
Balancing Chemical Equations
Balancing chemical equations is an essential skill in chemistry that ensures the law of conservation of mass is obeyed. According to this law, the number of each type of atom must be the same on both sides of a chemical equation.
To balance the combustion reaction of pentane and oxygen, one must first list all reactants and products. Begin with:\[\mathrm{C}_5\mathrm{H}_{12} + \mathrm{O}_2 \rightarrow \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O}\]
  • Balance carbon: Pentane has 5 carbons, so use 5 \(\mathrm{CO}_2\).
  • Balance hydrogen: Pentane has 12 hydrogens, so use 6 \(\mathrm{H}_2\mathrm{O}\).
  • Balance oxygen: Count total oxygens needed in products and adjust \(\mathrm{O}_2\) accordingly.
This step-by-step approach helps you achieve:\[\mathrm{C}_5\mathrm{H}_{12} + 8\mathrm{O}_2 \rightarrow 5\mathrm{CO}_2 + 6\mathrm{H}_2\mathrm{O}\] Understanding how to balance equations is key to mastering chemistry.
Oxidizing Agent
In a redox reaction, the oxidizing agent is the species that gets reduced. It facilitates oxidation by accepting electrons from another species. In our combustion reaction, oxygen (\(\mathrm{O}_2\)) takes on this role as it gains electrons from pentane. By doing so, it is converted to water (\(\mathrm{H}_2\mathrm{O}\)).
  • The oxidizing agent itself is reduced during the reaction.
  • This agent plays a crucial role in driving the reaction forward.
Understanding which species is the oxidizing agent is important for deciphering redox reaction mechanisms.
Reducing Agent
Conversely, the reducing agent in a redox reaction donates electrons, facilitating the reduction of another species. This means the reducing agent itself is oxidized. In the combustion of pentane, \(\mathrm{C}_5\mathrm{H}_{12}\) acts as the reducing agent. As it oxidizes, it loses electrons, transforming into carbon dioxide (\(\mathrm{CO}_2\)).
  • The reducing agent is oxidized and becomes a different chemical substance.
  • Recognizing the reducing agent helps to identify the electron flow in the reaction.
Identifying the reducing agent is just as crucial as finding the oxidizing agent as it completes the picture of electron transfer dynamics in the reaction.

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Most popular questions from this chapter

Ethyl chloride is prepared by the reaction of chlorine with ethane according to the following balanced equation. \(\mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}(\ell)+\mathrm{HCl}(\mathrm{g})\) When \(5.6 \mathrm{g}\) of ethane is reacted with excess chlorine, \(8.2 \mathrm{g}\) of ethyl chloride forms. Calculate the percent yield of ethyl chloride.

Calculate the number of moles in: (a) \(32 \mathrm{g}\) of methane, \(\mathrm{CH}_{4}\) (b) \(345.6 \mathrm{g}\) of nitric oxide, NO (c) \(184.4 \mathrm{g}\) of chlorine dioxide, \(\mathrm{ClO}_{2}\) (d) \(720 .\) g of glycerin, \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\)

Calculate the number of moles of: (a) \(\mathrm{S}^{2-}\) ions in \(6.56 \mathrm{mol}\) of \(\mathrm{Na}_{2} \mathrm{S}\) (b) \(\mathrm{Mg}^{2+}\) ions in \(8.320 \mathrm{mol}\) of \(\mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) (c) acetate ions, \(\mathrm{CH}_{3} \mathrm{COO}^{-},\) in \(0.43 \mathrm{mol}\) of \(\mathrm{Ca}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}\)

What is the mass in grams of each number of molecules of formaldehyde, \(\mathrm{CH}_{2} \mathrm{O} ?\) (a) \(100 .\) molecules (b) \(3000 .\) molecules (c) \(5.0 \times 10^{6}\) molecules (d) \(2.0 \times 10^{24}\) molecules

Answer true or false. (a) When a substance is oxidized, it loses electrons. (b) When a substance gains electrons, it is reduced. (c) In a redox reaction, the oxidizing agent becomes reduced. (d) In a redox reaction, the reducing reagent becomes oxidized. (e) When Zn is converted to \(\mathrm{Zn}^{2+}\) ion, zinc is oxidized. (f) Oxidation can also be defined as the loss of oxygen atoms and/or the gain of hydrogen atoms. (g) Reduction can also be defined as the gain of oxygen atoms and/or the loss of hydrogen atoms. (h) When oxygen, \(O_{2},\) is converted to hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2},\) we say that \(\mathrm{O}_{2}\) is reduced. (i) Hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\), is an oxidizing agent. (j) All combustion reactions are redox reactions. (k) The products of complete combustion (oxidation) of hydrocarbon fuels are carbon dioxide, water, and heat. (l) In the combustion of hydrocarbon fuels, oxygen is the oxidizing agent and the hydrocarbon fuel is the reducing agent. \((\mathrm{m})\) Incomplete combustion of hydrocarbon fuels can produce significant amounts of carbon monoxide. (n) Most common bleaches are oxidizing agents.
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