/*! 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 7 When solid carbon burns in a lim... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

When solid carbon burns in a limited supply of oxygen gas, the gas carbon monoxide, \(\mathrm{CO}\) forms. This gas is deadly to humans because it combines with hemoglobin in the blood, making it impossible for the blood to transport oxygen. Write a balanced equation for the formation of carbon monoxide.

Short Answer

Expert verified
The balanced equation is: \[ 2\mathrm{C} + \mathrm{O}_2 \rightarrow 2\mathrm{CO} \]

Step by step solution

01

Understanding the Basic Reaction

When carbon (C) reacts with oxygen (O2), they form carbon monoxide (CO). The chemical reaction can be summarized as: Carbon + Oxygen → Carbon Monoxide.
02

Writing the Unbalanced Equation

Based on the basic understanding of the reaction, we begin by writing the unbalanced chemical equation: \[ \mathrm{C} + \mathrm{O}_2 \rightarrow \mathrm{CO} \]
03

Counting and Comparing Atoms

In the unbalanced equation, we have 1 carbon (C) atom on the left and 1 carbon (C) atom on the right. Similarly, there are 2 oxygen (O) atoms on the left (from O2) and 1 oxygen (O) atom on the right.
04

Balancing the Oxygen Atoms

To balance the oxygen atoms, write: 2CO on the right side, which gives us 2 carbon monoxide molecules. This means adding a coefficient of 2 in front of CO: \[ \mathrm{C} + \mathrm{O}_2 \rightarrow 2\mathrm{CO} \]
05

Balancing the Carbon Atoms

Now, we notice that there are 2 carbon atoms needed on the right side. To balance this, add a coefficient of 2 in front of carbon: \[ 2\mathrm{C} + \mathrm{O}_2 \rightarrow 2\mathrm{CO} \]
06

Verification of Balance

We check both sides: there are 2 carbon atoms and 2 oxygen atoms on both sides. This verifies that the equation is balanced.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Balancing Chemical Equations
Balancing chemical equations is a crucial skill in chemistry, as it ensures that the same amount of matter is represented on both sides of a reaction. This is based on the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. Here’s how you can balance chemical equations effectively:

- **Start by writing the unbalanced equation**: Begin with the basic equation showing the reactants and products involved. For the combustion of carbon to form carbon monoxide, the unbalanced equation is \( \mathrm{C} + \mathrm{O}_2 \rightarrow \mathrm{CO} \).
- **Count the atoms**: List the number of atoms for each element on both sides of the equation. Initially, you will notice 1 carbon on each side but 2 oxygen atoms on the left and only 1 on the right.
- **Add coefficients to balance the atoms**: Adjust the equation by adding coefficients that make the number of each type of atom equal on both sides. For our example, placing a coefficient of 2 in front of \( \mathrm{CO} \) and \( \mathrm{C} \) balances the equation: \( 2\mathrm{C} + \mathrm{O}_2 \rightarrow 2\mathrm{CO} \).
- **Verify**: Always double-check your balanced equation to ensure that all atoms are accounted for equally on both sides. By adjusting coefficients—not changing subscripts—balances matter correctly without altering the compounds.
Carbon Monoxide Formation
Carbon monoxide ( CO) is produced in incomplete combustion processes, where there isn't enough oxygen for carbon to be completely oxidized to carbon dioxide ( CO 2). In our example, when solid carbon (charcoal or coal) combusts in limited oxygen, CO forms instead of CO 2. This process can be explained as follows:

- **Basic Reaction**: The reaction C + O 2 → CO occurs when the oxygen supply is insufficient.
- **Industrial and Domestic Relevance**: Besides occurring naturally, CO is significant in industrial processes and can become a hazard in homes if, for example, gas appliances are poorly ventilated.
- **Dangers**: Carbon monoxide is particularly dangerous because it is a colorless and odorless gas. It hinders the blood's oxygen transport capability, binding to hemoglobin much more effectively than oxygen, which can lead to serious health risks or even death.
Oxygen Limitation in Combustion
In combustion reactions, the presence and quantity of oxygen directly influence the type of products formed. When oxygen is limited, incomplete combustion occurs, which can have several implications:

- **Incomplete Combustion**: Incomplete combustion happens when there isn't enough oxygen to allow the fuel (like carbon) to react completely to form CO 2. Instead, carbon monoxide (CO) or even soot (carbon) is formed.
- **Environmental Impact**: The release of CO into the atmosphere is a concern due to its contribution to air pollution and its role in forming ground-level ozone, which can harm human health and the environment.
- **Energy Efficiency**: Incomplete combustion is inefficient, as it does not convert all the fuel's potential energy into useful energy, often resulting in wasted fuel and more emissions.
- **Preventive Measures**: Ensuring a sufficient oxygen supply and proper ventilation can help achieve complete combustion, reducing the adverse effects associated with carbon monoxide release.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Heats of reaction are frequently measured by monitoring the change in temperature of a water bath in which the reaction mixture is immersed. A water bath used for this purpose contains 2.000 L of water. In the course of the reaction, the temperature of the water rose \(4.85^{\circ} \mathrm{C}\) (a) How many calories were liberated by the reaction? (b) If \(2 \mathrm{kg}\) of a given reactant is consumed in the reaction, how many calories are liberated for each kilogram?

Calculate the number of moles of: (a) 0 atoms in 18.1 mol of formaldehyde, \(\mathrm{CH}_{2} \mathrm{O}\) (b) \(\mathrm{Br}\) atoms in \(0.41 \mathrm{mol}\) of bromoform, \(\mathrm{CHBr}_{3}\) (c) \(\quad O\) atoms in \(3.5 \times 10^{3}\) mol of \(A l_{2}\left(S O_{4}\right)_{3}\) (d) \(\mathrm{Hg}\) atoms in \(87 \mathrm{g}\) of \(\mathrm{HgO}\)

At one time, acetaldehyde was prepared industrially by the reaction of ethylene with air in the presence of a copper catalyst. How many grams of acetaldehyde can be prepared from 81.7 g of ethylene? \(2 \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \stackrel{\mathrm{Catal}_{5} \mathrm{s}}{\longrightarrow} 2 \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}(\mathrm{g})\)

Chlorophyll, the compound responsible for the green color of leaves and grasses, contains one atom of magnesium in each molecule. If the percentage by weight of magnesium in chlorophyll is \(2.72 \%,\) what is the molecular weight of chlorophyll?

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.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.