Calculating Water Mass From Methane Combustion A Stoichiometry Guide

In chemistry, stoichiometry plays a vital role in understanding the quantitative relationships between reactants and products in chemical reactions. Combustion reactions, a specific type of chemical process, involve the rapid reaction between a substance with an oxidant, usually oxygen, to produce heat and light. A classic example is the combustion of methane ($CH_4$), the primary component of natural gas. This article delves into the stoichiometric calculations required to determine the mass of water ($H_2O$) produced from the combustion of a given amount of methane. We will explore the balanced chemical equation for the reaction, the mole ratios between reactants and products, and the conversion of moles to grams using molar mass. This understanding is crucial in various fields, including chemical engineering, environmental science, and even everyday life applications like understanding the efficiency of fuel combustion in engines.

The balanced chemical equation for the combustion of methane is:

$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$

This equation signifies that one molecule of methane ($CH_4$) reacts with two molecules of oxygen ($O_2$) to produce one molecule of carbon dioxide ($CO_2$) and two molecules of water ($H_2O$). Crucially, this equation also represents the molar ratios involved in the reaction. It indicates that one mole of methane reacts with two moles of oxygen to yield one mole of carbon dioxide and two moles of water. This molar relationship is the cornerstone of stoichiometric calculations.

In this specific problem, we are given 2.00 moles of methane ($CH_4$). According to the balanced equation, for every one mole of methane combusted, two moles of water ($H_2O$) are produced. Therefore, the combustion of 2.00 moles of methane will produce:

2. 00 mol $CH_4$ * (2 mol $H_2O$ / 1 mol $CH_4$) = 4.00 mol $H_2O$

This calculation reveals that 4.00 moles of water are produced from the combustion of 2.00 moles of methane. However, the question asks for the mass of water produced, not the number of moles. To convert moles to grams, we need to use the molar mass of water.

The molar mass of a substance is the mass of one mole of that substance, typically expressed in grams per mole (g/mol). To calculate the molar mass of water ($H_2O$), we sum the atomic masses of its constituent atoms. The atomic mass of hydrogen (H) is approximately 1.01 g/mol, and the atomic mass of oxygen (O) is approximately 16.00 g/mol. Therefore, the molar mass of water is:

(2 * 1.01 g/mol H) + (1 * 16.00 g/mol O) = 18.02 g/mol $H_2O$

Now that we know the number of moles of water produced (4.00 mol) and the molar mass of water (18.02 g/mol), we can calculate the mass of water produced:

3. 00 mol $H_2O$ * 18.02 g/mol $H_2O$ = 72.08 g $H_2O$

Therefore, the combustion of 2.00 moles of methane produces approximately 72.08 grams of water. This result aligns closely with option C in the given choices.

The correct answer is C. 72.1 g. This problem demonstrates the fundamental principles of stoichiometry, specifically the use of balanced chemical equations to determine the quantitative relationships between reactants and products. By understanding mole ratios and molar masses, we can accurately predict the amount of product formed in a chemical reaction. This knowledge is essential for various applications, including chemical synthesis, industrial processes, and environmental analysis. Stoichiometric calculations allow chemists and engineers to optimize reactions, maximize product yield, and ensure efficient use of resources. In the context of methane combustion, understanding the amount of water produced is crucial for designing efficient combustion systems and managing water emissions.

While this problem focuses on the stoichiometric calculation of water produced from methane combustion, it's important to consider other factors that can influence the actual yield of water in a real-world scenario. These factors include:

• Incomplete Combustion: Incomplete combustion occurs when there is insufficient oxygen for the reaction to proceed to completion. This can lead to the formation of other products, such as carbon monoxide (CO) and soot (unburnt carbon), reducing the yield of water.
• Reaction Conditions: Temperature and pressure can affect the equilibrium of the reaction and the rate at which it proceeds. Optimal conditions are crucial for maximizing product yield.
• Purity of Reactants: Impurities in the methane or oxygen can interfere with the reaction and reduce the yield of water.
• Side Reactions: Other reactions may occur simultaneously, consuming reactants and reducing the yield of the desired product.

Despite these complexities, stoichiometric calculations provide a crucial theoretical framework for understanding chemical reactions. They allow us to predict the maximum amount of product that can be formed under ideal conditions. In real-world applications, these calculations serve as a starting point for optimizing reaction conditions and maximizing product yield.

The principles of stoichiometry are widely applied in various fields, including:

• Chemical Engineering: Stoichiometric calculations are essential for designing chemical reactors, optimizing reaction conditions, and ensuring efficient production of chemicals.
• Environmental Science: Understanding the stoichiometry of combustion reactions is crucial for assessing air pollution and developing strategies to reduce emissions of greenhouse gases and other pollutants.
• Pharmaceutical Industry: Stoichiometry is used to calculate the amounts of reactants needed to synthesize drugs and other pharmaceutical compounds.
• Materials Science: Stoichiometric calculations are used to design new materials with specific properties.
• Analytical Chemistry: Stoichiometry is used to determine the concentration of substances in samples through techniques such as titrations.

In summary, determining the mass of water produced from the combustion of methane involves understanding the balanced chemical equation, applying mole ratios, and converting moles to grams using molar mass. The stoichiometric calculation provides a theoretical yield, and other factors can influence the actual yield in real-world scenarios. The principles of stoichiometry are fundamental to chemistry and have wide-ranging applications in various scientific and industrial fields. By mastering these concepts, students and professionals can effectively analyze chemical reactions, predict product yields, and optimize chemical processes. The ability to perform stoichiometric calculations is a cornerstone of chemical literacy and is essential for anyone working in a field related to chemistry or chemical engineering. Understanding the quantitative relationships between reactants and products is crucial for making informed decisions about chemical processes and ensuring the efficient use of resources. Furthermore, the application of stoichiometry extends beyond the laboratory and into everyday life, informing our understanding of combustion processes, energy production, and environmental issues.

To further solidify your understanding of stoichiometry, consider working through the following practice problems:

1. What mass of carbon dioxide ($CO_2$) is produced by the combustion of 5.00 mol of methane ($CH_4$)?
2. If 100.0 g of oxygen ($O_2$) reacts with excess methane, what mass of water ($H_2O$) will be produced?
3. How many moles of methane are required to produce 36.0 g of water?

By solving these problems, you can reinforce your understanding of the concepts discussed in this article and improve your ability to apply stoichiometry to various chemical calculations.

For a deeper understanding of stoichiometry and related concepts, consider exploring the following resources:

• Textbooks on general chemistry and chemical principles
• Online chemistry courses and tutorials
• Scientific journals and publications
• Websites dedicated to chemistry education

By engaging with these resources, you can expand your knowledge of stoichiometry and its applications in chemistry and related fields.