Burning Propane Calculating Moles Of CO2 H2O And C3H8

Introduction

In the realm of chemistry, stoichiometry plays a pivotal role in understanding the quantitative relationships between reactants and products in chemical reactions. One common example is the combustion of hydrocarbons, where substances like propane react with oxygen to produce carbon dioxide and water. This article delves into a specific scenario: If 0.647 mol of oxygen is used in the burning of propane, we aim to determine how many moles of carbon dioxide (CO2) and water (H2O) are produced, as well as how many moles of propane (C3H8) are consumed. This exploration will not only provide the answers but also illuminate the fundamental principles underlying such calculations.

Understanding Stoichiometry and Combustion Reactions

Before diving into the specific problem, it's crucial to grasp the underlying concepts. Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. These relationships are governed by the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. This law is reflected in balanced chemical equations, which ensure that the number of atoms of each element is the same on both sides of the equation.

Combustion reactions, specifically, are exothermic chemical processes that involve the rapid reaction between a substance with an oxidant, usually oxygen, to produce heat and light. In the case of hydrocarbons like propane (C3H8), complete combustion results in the formation of carbon dioxide (CO2) and water (H2O). The general form of a combustion reaction for a hydrocarbon is:

CₓHᵧ + O₂ → CO₂ + H₂O

To solve stoichiometric problems related to combustion, the first step is to write and balance the chemical equation. This balanced equation provides the molar ratios between reactants and products, which are essential for calculating the amounts of substances involved in the reaction. These ratios act as conversion factors, allowing us to move from moles of one substance to moles of another.

Understanding these principles is foundational to solving the problem at hand. By correctly balancing the equation and interpreting the molar ratios, we can accurately determine the quantities of products formed and reactants consumed in the combustion of propane.

Balancing the Chemical Equation for Propane Combustion

The cornerstone of solving any stoichiometric problem lies in having a balanced chemical equation. For the combustion of propane (C3H8), the unbalanced equation is:

C3H8 + O2 → CO2 + H2O

To balance this equation, we need to ensure that the number of atoms for each element is the same on both the reactant and product sides. Let's break it down step by step:

1. Carbon (C): There are 3 carbon atoms in C3H8, so we need 3 CO2 molecules on the product side:

   C3H8 + O2 → 3 CO2 + H2O

2. Hydrogen (H): There are 8 hydrogen atoms in C3H8, so we need 4 H2O molecules on the product side:

   C3H8 + O2 → 3 CO2 + 4 H2O

3. Oxygen (O): Now, we count the oxygen atoms on the product side: 3 CO2 molecules have 3 * 2 = 6 oxygen atoms, and 4 H2O molecules have 4 * 1 = 4 oxygen atoms, totaling 10 oxygen atoms. To balance the oxygen, we need 5 O2 molecules on the reactant side:

   C3H8 + 5 O2 → 3 CO2 + 4 H2O

Now, the equation is balanced. This balanced equation is the key to unlocking the stoichiometric relationships in this reaction. It tells us that for every 1 mole of propane (C3H8) that reacts, 5 moles of oxygen (O2) are consumed, producing 3 moles of carbon dioxide (CO2) and 4 moles of water (H2O). This 1:5:3:4 molar ratio is what we will use to calculate the amounts of CO2, H2O, and C3H8 involved when 0.647 moles of oxygen are used.

Calculating Moles of CO2 and H2O Produced

With the balanced chemical equation in hand, we can now calculate the moles of carbon dioxide (CO2) and water (H2O) produced when 0.647 moles of oxygen (O2) are used in the combustion of propane. The balanced equation is:

C3H8 + 5 O2 → 3 CO2 + 4 H2O

This equation provides the molar ratios we need for our calculations. Specifically, it tells us that 5 moles of O2 produce 3 moles of CO2 and 4 moles of H2O. We can use these ratios as conversion factors to find the moles of CO2 and H2O produced from 0.647 moles of O2.

Calculating Moles of CO2

To find the moles of CO2 produced, we use the molar ratio between O2 and CO2, which is 5:3. This means that for every 5 moles of O2, 3 moles of CO2 are produced. We set up the calculation as follows:

Moles of CO2 = (0.647 moles O2) × (3 moles CO2 / 5 moles O2)

Moles of CO2 = 0.3882 moles

Rounding this to three significant figures, we get:

Moles of CO2 ≈ 0.388 moles

Calculating Moles of H2O

Similarly, to find the moles of H2O produced, we use the molar ratio between O2 and H2O, which is 5:4. This means that for every 5 moles of O2, 4 moles of H2O are produced. We set up the calculation as follows:

Moles of H2O = (0.647 moles O2) × (4 moles H2O / 5 moles O2)

Moles of H2O = 0.5176 moles

Rounding this to three significant figures, we get:

Moles of H2O ≈ 0.518 moles

Therefore, when 0.647 moles of oxygen are used in the burning of propane, approximately 0.388 moles of CO2 and 0.518 moles of H2O are produced. These calculations highlight the power of stoichiometry in predicting the outcomes of chemical reactions based on molar ratios.

Determining Moles of C3H8 Consumed

Having calculated the moles of CO2 and H2O produced, we now turn our attention to determining the moles of propane (C3H8) consumed in the reaction. Again, the balanced chemical equation is our guide:

C3H8 + 5 O2 → 3 CO2 + 4 H2O

This equation shows that 1 mole of propane (C3H8) reacts with 5 moles of oxygen (O2). We can use this molar ratio to calculate the amount of C3H8 consumed when 0.647 moles of O2 are used.

Calculating Moles of C3H8

The molar ratio between C3H8 and O2 is 1:5. This means that for every 5 moles of O2, 1 mole of C3H8 is consumed. We set up the calculation as follows:

Moles of C3H8 = (0.647 moles O2) × (1 mole C3H8 / 5 moles O2)

Moles of C3H8 = 0.1294 moles

Rounding this to three significant figures, we get:

Moles of C3H8 ≈ 0.129 moles

Therefore, when 0.647 moles of oxygen are used in the burning of propane, approximately 0.129 moles of propane are consumed. This calculation completes our understanding of the stoichiometry of this reaction, allowing us to quantify the amounts of all reactants and products involved.

Practical Implications and Real-World Applications

The calculations we've performed aren't just theoretical exercises; they have significant practical implications and real-world applications. Understanding the stoichiometry of combustion reactions is crucial in various fields, including:

1. Industrial Chemistry: In industrial processes, the precise control of chemical reactions is essential for efficiency and safety. Knowing the exact amounts of reactants required to produce a desired amount of product helps optimize processes and minimize waste. For example, in the production of plastics or pharmaceuticals, stoichiometric calculations ensure that reactions proceed as expected.

2. Environmental Science: Combustion processes are a major source of air pollution. By understanding the stoichiometry of these reactions, we can develop strategies to reduce the emission of harmful pollutants. For instance, optimizing the air-to-fuel ratio in combustion engines can minimize the formation of carbon monoxide (CO), a toxic gas.

3. Engineering: Engineers use stoichiometric principles in the design and operation of combustion engines, power plants, and other systems involving chemical reactions. Accurate calculations are necessary to ensure that these systems function efficiently and safely. For example, in designing a power plant, engineers need to know the exact amount of fuel required to generate a certain amount of electricity.

4. Safety: Understanding stoichiometry is crucial for preventing and mitigating chemical accidents. Knowing the potential hazards associated with different reactions and the amounts of reactants that can lead to explosions or fires is essential for laboratory and industrial safety.

5. Everyday Life: Even in everyday life, stoichiometric principles are at play. Cooking, for instance, involves chemical reactions, and following recipes often requires precise measurements of ingredients to achieve the desired outcome. Similarly, understanding the combustion of fuels in heating systems helps us use energy efficiently.

The ability to perform stoichiometric calculations, as demonstrated in this article, is a fundamental skill in chemistry and related fields. It allows us to predict the outcomes of chemical reactions, optimize industrial processes, protect the environment, and ensure safety in various applications.

Conclusion

In this comprehensive exploration, we addressed the question: If 0.647 mol of oxygen is used in the burning of propane, how many moles each of CO2 and H2O are produced, and how many moles of C3H8 are consumed? Through a step-by-step approach, we first balanced the chemical equation for the combustion of propane:

C3H8 + 5 O2 → 3 CO2 + 4 H2O

Using this balanced equation, we then calculated the moles of carbon dioxide (CO2) and water (H2O) produced, as well as the moles of propane (C3H8) consumed:

• Moles of CO2 produced: approximately 0.388 moles
• Moles of H2O produced: approximately 0.518 moles
• Moles of C3H8 consumed: approximately 0.129 moles

These calculations demonstrate the power of stoichiometry in quantifying the relationships between reactants and products in chemical reactions. We also discussed the practical implications and real-world applications of stoichiometric principles in various fields, including industrial chemistry, environmental science, engineering, safety, and everyday life.

Understanding stoichiometry is not just an academic exercise; it is a crucial skill for anyone working with chemical reactions. It allows us to predict outcomes, optimize processes, and ensure safety. By mastering these fundamental concepts, we can better understand and interact with the world around us.

This article serves as a testament to the importance of stoichiometry and provides a clear, step-by-step guide to solving similar problems. Whether you are a student learning chemistry or a professional working in a related field, the principles and calculations discussed here will undoubtedly prove valuable in your endeavors.