The Equation of Sucrose Combustion

Oct 13, 2025 by wbrum.org 16 views

Introduction

Chemical reactions are the foundation of chemistry, and understanding the equations that represent these reactions is crucial for analyzing and predicting the outcomes of various processes. In this article, we will delve into the combustion of sucrose, a common sugar found in many foods, and explore the equation that represents this reaction. We will also use this equation to solve a problem involving the combustion of sucrose and oxygen.

The Equation of Sucrose Combustion

The equation for the combustion of sucrose is given by:

${ C_{12} H_{22} O_{11} + 12 O_2 \rightarrow 12 CO_2 + 11 H_2 O }$

This equation indicates that when sucrose reacts with oxygen, it produces carbon dioxide and water. The coefficients in the equation represent the number of moles of each substance involved in the reaction.

Understanding the Coefficients

The coefficients in the equation represent the number of moles of each substance involved in the reaction. In this case, the coefficient of sucrose is 1, indicating that 1 mole of sucrose reacts with 12 moles of oxygen to produce 12 moles of carbon dioxide and 11 moles of water.

Solving the Problem

Now that we have a clear understanding of the equation, let's use it to solve a problem. Suppose we have 10.0 g of sucrose and 8.0 g of oxygen, and we want to know how many moles of carbon dioxide and water will be produced.

Step 1: Calculate the Number of Moles of Sucrose

To calculate the number of moles of sucrose, we need to know its molar mass. The molar mass of sucrose is 342.3 g/mol. We can calculate the number of moles of sucrose using the following formula:

${ \text{moles} = \frac{\text{mass}}{\text{molar mass}} }$

Plugging in the values, we get:

${ \text{moles of sucrose} = \frac{10.0 \text{ g}}{342.3 \text{ g/mol}} = 0.0291 \text{ mol} }$

Step 2: Calculate the Number of Moles of Oxygen

To calculate the number of moles of oxygen, we need to know its molar mass. The molar mass of oxygen is 32.0 g/mol. We can calculate the number of moles of oxygen using the following formula:

${ \text{moles} = \frac{\text{mass}}{\text{molar mass}} }$

Plugging in the values, we get:

${ \text{moles of oxygen} = \frac{8.0 \text{ g}}{32.0 \text{ g/mol}} = 0.25 \text{ mol} }$

Step 3: Calculate the Number of Moles of Carbon Dioxide and Water

Now that we have the number of moles of sucrose and oxygen, we can calculate the number of moles of carbon dioxide and water produced using the coefficients in the equation.

The coefficient of carbon dioxide is 12, so the number of moles of carbon dioxide produced is:

${ \text{moles of carbon dioxide} = 12 \times 0.0291 \text{ mol} = 0.3492 \text{ mol} }$

The coefficient of water is 11, so the number of moles of water produced is:

${ \text{moles of water} = 11 \times 0.0291 \text{ mol} = 0.3191 \text{ mol} }$

Conclusion

In this article, we have explored the equation that represents the combustion of sucrose and used it to solve a problem involving the combustion of sucrose and oxygen. We have calculated the number of moles of carbon dioxide and water produced using the coefficients in the equation. This problem demonstrates the importance of understanding chemical equations and how they can be used to analyze and predict the outcomes of various processes.

The Importance of Chemical Equations

Chemical equations are a fundamental concept in chemistry, and understanding them is crucial for analyzing and predicting the outcomes of various processes. Chemical equations provide a concise and accurate representation of chemical reactions, allowing us to identify the reactants and products involved, as well as the conditions under which the reaction occurs.

The Role of Coefficients in Chemical Equations

Coefficients in chemical equations represent the number of moles of each substance involved in the reaction. They are essential for understanding the stoichiometry of a reaction, which is the relationship between the amounts of reactants and products involved.

The Significance of Molar Mass

Molar mass is a critical concept in chemistry, as it allows us to calculate the number of moles of a substance using its mass. In this article, we have used the molar mass of sucrose and oxygen to calculate the number of moles of each substance involved in the reaction.

The Importance of Stoichiometry

Stoichiometry is the study of the quantitative relationships between reactants and products in chemical reactions. It is a critical concept in chemistry, as it allows us to predict the outcomes of various processes and understand the conditions under which a reaction occurs.

The Role of Chemical Equations in Real-World Applications

Chemical equations have numerous real-world applications, including the development of new materials, the design of chemical processes, and the analysis of environmental systems. In this article, we have demonstrated the importance of chemical equations in solving a problem involving the combustion of sucrose and oxygen.

Conclusion

Q: What is the equation for the combustion of sucrose?

A: The equation for the combustion of sucrose is:

${ C_{12} H_{22} O_{11} + 12 O_2 \rightarrow 12 CO_2 + 11 H_2 O }$

Q: What is the role of oxygen in the combustion of sucrose?

A: Oxygen is a reactant in the combustion of sucrose, and it plays a crucial role in the reaction. It is necessary for the sucrose to undergo combustion and produce carbon dioxide and water.

Q: What is the product of the combustion of sucrose?

A: The products of the combustion of sucrose are carbon dioxide and water. The carbon dioxide is produced in a 1:1 ratio with the sucrose, while the water is produced in a 1:11 ratio with the sucrose.

Q: How many moles of carbon dioxide are produced when 1 mole of sucrose is combusted?

A: According to the equation, 1 mole of sucrose produces 12 moles of carbon dioxide.

Q: How many moles of water are produced when 1 mole of sucrose is combusted?

A: According to the equation, 1 mole of sucrose produces 11 moles of water.

Q: What is the molar mass of sucrose?

A: The molar mass of sucrose is 342.3 g/mol.

Q: What is the molar mass of oxygen?

A: The molar mass of oxygen is 32.0 g/mol.

Q: How many moles of sucrose are present in 10.0 g of sucrose?

A: To calculate the number of moles of sucrose, we can use the following formula:

${ \text{moles} = \frac{\text{mass}}{\text{molar mass}} }$

Plugging in the values, we get:

${ \text{moles of sucrose} = \frac{10.0 \text{ g}}{342.3 \text{ g/mol}} = 0.0291 \text{ mol} }$

Q: How many moles of oxygen are present in 8.0 g of oxygen?

A: To calculate the number of moles of oxygen, we can use the following formula:

${ \text{moles} = \frac{\text{mass}}{\text{molar mass}} }$

Plugging in the values, we get:

${ \text{moles of oxygen} = \frac{8.0 \text{ g}}{32.0 \text{ g/mol}} = 0.25 \text{ mol} }$

Q: What is the relationship between the number of moles of sucrose and oxygen in the combustion reaction?

A: According to the equation, 1 mole of sucrose requires 12 moles of oxygen to undergo combustion.

Q: What is the significance of the coefficients in the equation for the combustion of sucrose?

A: The coefficients in the equation represent the number of moles of each substance involved in the reaction. They are essential for understanding the stoichiometry of the reaction.

Q: What is the importance of understanding the equation for the combustion of sucrose?

A: Understanding the equation for the combustion of sucrose is crucial for analyzing and predicting the outcomes of various processes. It allows us to identify the reactants and products involved, as well as the conditions under which the reaction occurs.

Q: What are some real-world applications of the equation for the combustion of sucrose?

A: The equation for the combustion of sucrose has numerous real-world applications, including the development of new materials, the design of chemical processes, and the analysis of environmental systems.