Calculating Oxygen Volume For Ethene Combustion Stoichiometry Explained

Jul 5, 2025 by Admin 72 views

Stoichiometry of Ethene Combustion How Much Oxygen is Needed?

In the realm of chemistry, stoichiometry serves as a fundamental pillar, providing the tools to unravel the quantitative relationships within chemical reactions. Among the myriad applications of stoichiometry, one stands out for its practical significance: determining the precise amounts of reactants and products involved in a given reaction. This exploration delves into the fascinating world of ethene combustion, employing stoichiometry to calculate the volume of oxygen required to react completely with a specific quantity of ethene at standard temperature and pressure (STP).

Stoichiometry Unveiled The Art of Chemical Calculations

At its core, stoichiometry is the language of chemical proportions, enabling us to predict the outcomes of chemical reactions with remarkable accuracy. It empowers us to answer critical questions such as: How much reactant is needed to produce a desired amount of product? Or, conversely, how much product can be generated from a given quantity of reactant? To navigate this intricate landscape, we rely on balanced chemical equations, which serve as roadmaps, guiding us through the quantitative relationships between reactants and products.

Balanced chemical equations are more than just symbolic representations of chemical reactions; they are quantitative statements, revealing the molar ratios in which substances react and are formed. These ratios are the cornerstone of stoichiometric calculations, allowing us to convert between the amounts of different species involved in the reaction. To illustrate, let's consider the balanced equation for the combustion of ethene:

C_2 H_{4(g)} + 3 O_{2(g)} \rightarrow 2 CO_{2(g)} + 2 H_2 O_{(g)}

This equation reveals that one mole of ethene ( $C_2H_4$ ) reacts with three moles of oxygen ( $O_2$ ) to produce two moles of carbon dioxide ( $CO_2$ ) and two moles of water ( $H_2O$ ). These molar ratios are the key to unlocking stoichiometric calculations.

Ethene Combustion A Stoichiometric Puzzle

The combustion of ethene is a classic example of a chemical reaction that lends itself perfectly to stoichiometric analysis. Ethene, a gaseous hydrocarbon, readily reacts with oxygen in a fiery dance, producing carbon dioxide and water as the main products. This reaction is not merely a chemical curiosity; it has practical implications, serving as the foundation for various industrial processes and combustion technologies.

The balanced equation for ethene combustion provides the stoichiometric framework for our calculations. It tells us that for every one molecule of ethene that participates in the reaction, three molecules of oxygen are required to ensure complete combustion. This 1:3 molar ratio between ethene and oxygen is the crux of our calculation.

Volume Calculation Unveiling the Quantitative Relationship

Our challenge is to determine the volume of oxygen needed to react with 270 liters of ethene at STP. To tackle this, we'll leverage the concept of molar volume, which states that one mole of any gas occupies 22.4 liters at STP. This principle allows us to seamlessly convert between volumes and moles, bridging the gap between macroscopic measurements and the microscopic world of molecules.

Here's the step-by-step approach to solving this problem:

Convert the volume of ethene to moles. At STP, 22.4 liters of any gas corresponds to one mole. Therefore, 270 liters of ethene is equivalent to:
$Moles ext{ } of ext{ } ethene = rac{270 ext{ } liters}{22.4 ext{ } liters/mole} = 12.05 ext{ } moles$
Apply the stoichiometric ratio. From the balanced equation, we know that 1 mole of ethene reacts with 3 moles of oxygen. So, 12.05 moles of ethene will react with:
$Moles ext{ } of ext{ } oxygen = 12.05 ext{ } moles ext{ } ethene imes rac{3 ext{ } moles ext{ } oxygen}{1 ext{ } mole ext{ } ethene} = 36.15 ext{ } moles$
Convert moles of oxygen back to volume. Using the molar volume concept, we can convert 36.15 moles of oxygen to volume:
$Volume ext{ } of ext{ } oxygen = 36.15 ext{ } moles imes 22.4 ext{ } liters/mole = 809.76 ext{ } liters$

Therefore, approximately 809.76 liters of oxygen are required to react completely with 270 liters of ethene at STP.

The Correct Answer A Stoichiometric Triumph

Through the power of stoichiometry, we've successfully determined the volume of oxygen required for the complete combustion of ethene. The correct answer is approximately 809.76 liters. This exercise not only showcases the practical application of stoichiometry but also underscores the importance of balanced chemical equations in unraveling quantitative relationships in chemical reactions.

Stoichiometry Beyond Combustion A Versatile Tool

The principles of stoichiometry extend far beyond combustion reactions. They are applicable to a vast array of chemical processes, including acid-base reactions, precipitation reactions, and redox reactions. Stoichiometry is the bedrock of chemical calculations, enabling us to design experiments, optimize industrial processes, and understand the fundamental interactions between matter at the molecular level.

Mastering Stoichiometry A Journey of Chemical Discovery

Stoichiometry is not merely a set of formulas and calculations; it's a way of thinking about chemical reactions. By mastering stoichiometry, we gain a deeper appreciation for the quantitative nature of chemistry and the elegance with which chemical reactions unfold. It's a journey of discovery that unlocks the secrets of the molecular world.

In the world of chemistry, stoichiometry serves as a powerful tool for quantifying the relationships between reactants and products in chemical reactions. It allows us to predict the amount of substances consumed or produced in a reaction, making it essential for various applications, from industrial processes to laboratory experiments. This discussion delves into a specific application of stoichiometry: calculating the volume of oxygen required to react with a given amount of ethene ( $C_2H_4$ ) at standard temperature and pressure (STP).

Understanding Stoichiometry The Foundation of Chemical Calculations

At its core, stoichiometry is the study of quantitative relationships in chemical reactions. It relies on the fundamental principle that matter is neither created nor destroyed in a chemical reaction, meaning the number of atoms of each element must be the same on both sides of a balanced chemical equation. These balanced equations provide the crucial information needed for stoichiometric calculations, acting as a roadmap to navigate the quantitative aspects of chemical reactions.

A balanced chemical equation not only represents the chemical transformation but also reveals the molar ratios in which reactants and products participate. For instance, consider the balanced equation for the combustion of ethene:

C_2H_{4(g)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 2H_2O_{(g)}

This equation signifies that one mole of ethene reacts with three moles of oxygen to produce two moles of carbon dioxide and two moles of water. These molar ratios are the key to performing stoichiometric calculations, allowing us to convert between the amounts of different substances involved in the reaction.

The Ethene Combustion Scenario A Stoichiometric Problem

The combustion of ethene is a classic example of a chemical reaction with significant industrial applications. Ethene, a simple alkene, readily reacts with oxygen, releasing a substantial amount of energy in the form of heat and light. This reaction is the basis for many industrial processes, including the production of polymers and the generation of electricity.

In this scenario, we aim to determine the volume of oxygen required to completely react with 270 liters of ethene at STP. This problem exemplifies the practical application of stoichiometry, demonstrating how we can use balanced chemical equations and molar relationships to solve real-world chemical problems.

Calculation Steps A Stoichiometric Journey

To solve this problem, we will employ a step-by-step approach, utilizing the principles of stoichiometry and the concept of molar volume. At STP, one mole of any gas occupies 22.4 liters, a crucial conversion factor for relating volumes to moles.

Convert the volume of ethene to moles. Using the molar volume concept, we can convert the given volume of ethene (270 liters) to moles:
$Moles ext{ } of ext{ } ethene = rac{270 ext{ } liters}{22.4 ext{ } liters/mole} = 12.05 ext{ } moles$
Apply the stoichiometric ratio. From the balanced chemical equation, we know that one mole of ethene reacts with three moles of oxygen. Therefore, to react with 12.05 moles of ethene, we need:
$Moles ext{ } of ext{ } oxygen = 12.05 ext{ } moles ext{ } ethene imes rac{3 ext{ } moles ext{ } oxygen}{1 ext{ } mole ext{ } ethene} = 36.15 ext{ } moles$
Convert moles of oxygen to volume. Again, using the molar volume concept, we can convert the moles of oxygen (36.15 moles) back to volume:
$Volume ext{ } of ext{ } oxygen = 36.15 ext{ } moles imes 22.4 ext{ } liters/mole = 809.76 ext{ } liters$

Therefore, approximately 809.76 liters of oxygen are required to completely react with 270 liters of ethene at STP. This calculation demonstrates the power of stoichiometry in predicting the quantitative aspects of chemical reactions.

Conclusion Stoichiometry The Key to Quantitative Chemistry

Through this exercise, we have successfully calculated the volume of oxygen needed for the complete combustion of ethene. The result, approximately 809.76 liters, highlights the importance of stoichiometry in determining the precise amounts of reactants and products involved in chemical reactions.

Stoichiometry is not just a theoretical concept; it has practical implications across various fields, including chemical engineering, environmental science, and materials science. By mastering stoichiometric principles, we gain a deeper understanding of chemical reactions and their quantitative nature, enabling us to solve real-world problems and advance scientific knowledge.

In chemistry, stoichiometry is the branch that deals with the quantitative relationships between reactants and products in chemical reactions. It provides a framework for calculating the amounts of substances involved in chemical reactions, making it an essential tool for chemists and researchers. This discussion focuses on applying stoichiometry to determine the volume of oxygen required to react with a specific amount of ethene ( $C_2H_4$ ) at standard temperature and pressure (STP).

Unveiling Stoichiometry The Art of Balancing Equations

Stoichiometry is rooted in the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. This principle translates into the requirement that a balanced chemical equation must have the same number of atoms of each element on both sides of the equation. Balancing chemical equations is the first step in any stoichiometric calculation, ensuring that we have an accurate representation of the molar relationships between reactants and products.

The balanced chemical equation for the combustion of ethene is:

C_2H_{4(g)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 2H_2O_{(g)}

This equation tells us that one mole of ethene reacts with three moles of oxygen to produce two moles of carbon dioxide and two moles of water. These molar ratios are the foundation for all subsequent stoichiometric calculations.

Problem Statement Determining Oxygen Volume at STP

Our goal is to calculate the volume of oxygen required to react completely with 270 liters of ethene at STP. This problem combines stoichiometric principles with the concept of molar volume. At STP, one mole of any gas occupies 22.4 liters, a crucial piece of information that allows us to convert between volume and moles.

Solution Methodology A Step-by-Step Approach

To solve this problem, we will follow a systematic approach, breaking down the calculation into manageable steps:

Convert the volume of ethene to moles. We begin by converting the given volume of ethene (270 liters) to moles using the molar volume concept:
$Moles ext{ } of ext{ } ethene = rac{270 ext{ } liters}{22.4 ext{ } liters/mole} = 12.05 ext{ } moles$
Apply the stoichiometric ratio. From the balanced chemical equation, we know that one mole of ethene reacts with three moles of oxygen. Therefore, to react with 12.05 moles of ethene, we need:
$Moles ext{ } of ext{ } oxygen = 12.05 ext{ } moles ext{ } ethene imes rac{3 ext{ } moles ext{ } oxygen}{1 ext{ } mole ext{ } ethene} = 36.15 ext{ } moles$
Convert moles of oxygen to volume. Finally, we convert the moles of oxygen (36.15 moles) back to volume using the molar volume concept:
$Volume ext{ } of ext{ } oxygen = 36.15 ext{ } moles imes 22.4 ext{ } liters/mole = 809.76 ext{ } liters$

Therefore, approximately 809.76 liters of oxygen are required to completely react with 270 liters of ethene at STP. This result demonstrates the power of stoichiometry in quantifying chemical reactions.

Significance of Stoichiometry Beyond Calculations

Stoichiometry is more than just a set of calculations; it's a fundamental principle that underpins many aspects of chemistry. It allows us to predict the outcomes of chemical reactions, optimize reaction conditions, and design experiments with precision. Stoichiometry is essential for various applications, including industrial chemistry, environmental science, and pharmaceutical development.

By mastering stoichiometric principles, we gain a deeper understanding of the quantitative nature of chemistry and its applications in the real world. It's a skill that empowers us to solve chemical problems and contribute to scientific advancements.

Stoichiometry, the quantitative study of chemical reactions, provides the tools to understand the relationships between reactants and products. One practical application of stoichiometry is determining the amount of reactants needed for a complete reaction. This guide focuses on calculating the volume of oxygen required to react with 270 liters of ethene ( $C_2H_4$ ) at standard temperature and pressure (STP).

Stoichiometry Fundamentals Balanced Equations and Molar Ratios

Stoichiometry hinges on the law of conservation of mass, ensuring that atoms are neither created nor destroyed in chemical reactions. This principle is reflected in balanced chemical equations, where the number of atoms of each element is equal on both sides. Balanced equations reveal molar ratios, which are crucial for stoichiometric calculations.

For the combustion of ethene, the balanced equation is:

C_2H_{4(g)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 2H_2O_{(g)}

This equation indicates that one mole of ethene reacts with three moles of oxygen. This 1:3 molar ratio is the foundation for calculating the oxygen volume.

Problem Analysis Oxygen Requirement for Ethene Combustion

The problem asks for the volume of oxygen needed to completely react with 270 liters of ethene at STP. To solve this, we'll use stoichiometry and the molar volume concept. At STP, one mole of any gas occupies 22.4 liters, providing a bridge between volume and moles.

Step-by-Step Solution A Stoichiometric Walkthrough

Here's the detailed calculation process:

Convert ethene volume to moles. Using the molar volume at STP:
$Moles ext{ } of ext{ } ethene = rac{270 ext{ } liters}{22.4 ext{ } liters/mole} = 12.05 ext{ } moles$
Apply the ethene-to-oxygen molar ratio. From the balanced equation, 1 mole of ethene reacts with 3 moles of oxygen:
$Moles ext{ } of ext{ } oxygen = 12.05 ext{ } moles ext{ } ethene imes rac{3 ext{ } moles ext{ } oxygen}{1 ext{ } mole ext{ } ethene} = 36.15 ext{ } moles$
Convert oxygen moles to volume. Using the molar volume at STP:
$Volume ext{ } of ext{ } oxygen = 36.15 ext{ } moles imes 22.4 ext{ } liters/mole = 809.76 ext{ } liters$

Therefore, approximately 809.76 liters of oxygen are required to react completely with 270 liters of ethene at STP.

Stoichiometry's Broader Role Chemical Calculations and Beyond

Stoichiometry is a cornerstone of chemistry, essential for calculating reactant and product quantities. It is applied in various fields, including chemical synthesis, industrial processes, and environmental chemistry. Understanding stoichiometry allows chemists to predict reaction outcomes and optimize chemical processes.

By mastering stoichiometry, students and professionals gain a deeper insight into chemical reactions and their practical applications. This knowledge is crucial for advancing scientific understanding and developing new technologies.

In chemical reactions, stoichiometry plays a vital role in determining the quantitative relationships between reactants and products. A classic example is calculating the amount of oxygen required for the complete combustion of a given amount of ethene ( $C_2H_4$ ). This problem showcases how stoichiometric principles and molar relationships can be used to solve real-world chemistry questions.

Stoichiometry Essentials Balanced Equations and Molar Relationships

Stoichiometry is based on the law of conservation of mass, where atoms are neither created nor destroyed during a chemical reaction. Balanced chemical equations reflect this law, with an equal number of each type of atom on both sides. These equations reveal molar ratios, the foundation for stoichiometric calculations.

The balanced equation for ethene combustion is:

C_2H_{4(g)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 2H_2O_{(g)}

This equation shows that one mole of ethene reacts with three moles of oxygen, a critical 1:3 molar ratio for calculations.

Problem Definition Calculating Oxygen Volume at STP

The problem is to find the volume of oxygen needed to react completely with 270 liters of ethene at standard temperature and pressure (STP). We'll use stoichiometry and the molar volume concept, where one mole of any gas occupies 22.4 liters at STP.

Step-by-Step Calculation A Stoichiometry Solution

Here's a step-by-step approach to solve the problem:

Convert ethene volume to moles. Using the molar volume at STP:
$Moles ext{ } of ext{ } ethene = rac{270 ext{ } liters}{22.4 ext{ } liters/mole} = 12.05 ext{ } moles$
Apply the molar ratio from the balanced equation. For ethene and oxygen, the ratio is 1:3:
$Moles ext{ } of ext{ } oxygen = 12.05 ext{ } moles ext{ } ethene imes rac{3 ext{ } moles ext{ } oxygen}{1 ext{ } mole ext{ } ethene} = 36.15 ext{ } moles$
Convert oxygen moles to volume at STP. Using the molar volume:
$Volume ext{ } of ext{ } oxygen = 36.15 ext{ } moles imes 22.4 ext{ } liters/mole = 809.76 ext{ } liters$

Therefore, approximately 809.76 liters of oxygen are needed to react completely with 270 liters of ethene at STP.

Stoichiometry's Importance Practical Applications and Insights

Stoichiometry is a fundamental concept in chemistry, used to determine the quantitative aspects of chemical reactions. Its applications range from industrial chemical production to environmental analysis and research. Mastering stoichiometry allows for accurate predictions of reaction outcomes and optimization of chemical processes.

By understanding and applying stoichiometric principles, chemists and scientists can tackle complex problems and develop innovative solutions in various scientific and industrial fields.