Question

Imagine you mix \(16.05 \mathrm{~g}\) of methane \(\left(\mathrm{CH}_{4}\right)\) gas and \(96.00 \mathrm{~g}\) of oxygen \(\left(\mathrm{O}_{2}\right)\) gas and then ignite the mixture. After a bright flash and a loud bang, some water vapor forms. a. Write the balanced chemical reaction for the combustion of methane. b. Depict the process that occurred using circles to represent atoms. Represent carbon with black circles, hydrogen with white circles, and oxygen with gray circles. Let one circle (or one molecule made of circles bonded together) represent exactly one mole. c. How many moles of water can you make? How many moles of carbon dioxide? d. Will anything be left over? If so, how much? e. Identify the following: limiting reagent, excess reagent, and theoretical yield.

Short answer

The balanced chemical equation is \(\mathrm{CH}_4 + 2\mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O}\). 1.00 mole of methane reacts with 2.00 moles of oxygen to produce 2.00 moles of water and 1.00 mole of carbon dioxide, with 1.00 mole of oxygen left over. Methane is the limiting reagent, oxygen is the excess reagent, and the theoretical yield is 2.00 moles of water and 1.00 mole of carbon dioxide.

Step-by-step solution

Write the balanced chemical equation for methane combustion

The unbalanced chemical equation for the combustion of methane with oxygen is: \( \mathrm{CH}_4 + \mathrm{O}_2 \rightarrow \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O} \). To balance the equation, adjust the coefficients to ensure the same number of atoms of each element on both sides of the equation: \( \mathrm{CH}_4 + 2\mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O} \). This balanced equation indicates that 1 mole of methane reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 moles of water.

Depict the process using circles to represent atoms

Represent carbon with a black circle, hydrogen with white circles, and oxygen with gray circles. One molecule of methane (\(\mathrm{CH}_4\)) would be shown as one black circle bonded to four white circles. One molecule of oxygen (\(\mathrm{O}_2\)) is represented by two gray circles bonded together. After combustion, one molecule of carbon dioxide (\(\mathrm{CO}_2\)) would be one black circle bonded to two gray circles, and one molecule of water (\(\mathrm{H}_2\mathrm{O}\)) would be two white circles bonded to a gray circle.

Calculate the moles of methane and oxygen

Using the molecular weights (\(16.05~\mathrm{g}\) of methane with a molar mass of \(16.04~\mathrm{g/mol}\) and \(96.00~\mathrm{g}\) of oxygen with a molar mass of \(32.00~\mathrm{g/mol}\)), calculate the number of moles for each: \[ \text{Moles of }\mathrm{CH}_4 = \frac{16.05~\mathrm{g}}{16.04~\mathrm{g/mol}} \approx 1.00~\mathrm{mol} \] \[ \text{Moles of }\mathrm{O}_2 = \frac{96.00~\mathrm{g}}{32.00~\mathrm{g/mol}} = 3.00~\mathrm{mol} \]

Determine the limiting reagent and calculate the moles of products

According to the balanced equation, 2 moles of \(\mathrm{O}_2\) are required for every mole of \(\mathrm{CH}_4\). Therefore, methane is the limiting reagent because there is more than enough oxygen to react with the methane available. The theoretical yield of water will be twice the moles of methane, and the yield of carbon dioxide will be equal to the moles of methane: \[ \text{Moles of }\mathrm{H}_2\mathrm{O} = 2 \times 1.00~\mathrm{mol} = 2.00~\mathrm{mol} \] \[ \text{Moles of }\mathrm{CO}_2 = 1.00~\mathrm{mol} \]

Calculate any excess reactants

Find out if any oxygen is left over after the reaction: \( 1.00~\mathrm{mol}~\mathrm{CH}_4 \) uses up \( 2.00~\mathrm{mol}~\mathrm{O}_2 \), out of the 3.00 moles available. Thus, \[ \text{Excess }\mathrm{O}_2 = 3.00~\mathrm{mol} - 2.00~\mathrm{mol} = 1.00~\mathrm{mol} \] There will be 1.00 mole of oxygen left over.

Identify the limiting reagent, excess reagent, and theoretical yield

The limiting reagent is methane (\(\mathrm{CH}_4\)), as it determines the maximum amount of product that can be formed. The excess reagent is oxygen (\(\mathrm{O}_2\)), with 1.00 mole remaining unused. The theoretical yield is 2.00 moles of water and 1.00 mole of carbon dioxide, which are the maximum amounts that can be produced from the limiting reagent.

Key concepts

Balanced Chemical Equation

Understanding the balanced chemical equation is crucial in grasping how reactions work. In a balanced equation, the number of atoms for each element in the reactants side is equal to the number of atoms in the products side. For the combustion of methane, the equation is balanced as: \( \mathrm{CH}_4 + 2\mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O} \). This represents a 1:2:1:2 ratio respectively, ensuring that the law of conservation of mass is obeyed. Thus, through balancing, we determine the exact proportions of reactants needed to form products.

A balanced equation allows us to predict how much of each substance will be involved and created. It's the foundation for all subsequent steps in analyzing a chemical reaction, including stoichiometry, which calculates the quantitative relationships and conversions between reactants and products.

Limiting Reagent

The limiting reagent is the substance in a chemical reaction that runs out first, thus determining the amount of product that can be formed. In our methane combustion example, methane (\(\mathrm{CH}_4\)) is the limiting reagent. Even if there's plenty of oxygen, the reaction stops when all methane is consumed. This concept is a linchpin in chemical reactions because it allows us to calculate the maximum amount of product that can be produced, a value known as the theoretical yield.

Understanding which reactant is the limiting reagent helps prevent the wastage of materials in industrial processes and is essential for lab experiments where precise outcomes are needed.

Theoretical Yield

The theoretical yield is the maximum amount of product that can be produced in a chemical reaction based on the limiting reagent. It's a vital concept because it sets the benchmark for what we can expect if everything goes perfectly, with no side reactions or losses. The calculations from our methane combustion indicate that the theoretical yield from 1 mole of methane is 2 moles of water and 1 mole of carbon dioxide.

Real-life reactions often yield less than the theoretical yield due to various factors like inefficient reactions or loss of material, making this an ideal, best-case scenario for scientists and engineers to measure efficiency and for students to understand the stoichiometric relationship between reactants and products.

Mole Concept

The mole concept is a bridge between the microscopic world of atoms and molecules and the macroscopic world we measure in the lab. A mole represents Avogadro's number (\(6.022 \times 10^{23}\) entities) and allows chemists to count particles by weighing them. In the combustion exercise, we see that 16.05 grams of methane corresponds to roughly 1 mole, based on its molar mass of 16.04 grams per mole.

With the mole concept, it's possible to perform calculations involving mass and number of particles, facilitating the transition from the abstract formulae of substances to tangible quantities that can be manipulated and measured.

Stoichiometry

Stoichiometry is the field of chemistry that involves calculating the relative quantities of reactants and products in chemical reactions. It is based on the balanced chemical equations and the mole concept. In our combustion example, stoichiometry tells us how many moles of oxygen are needed to react with methane and how many moles of water and carbon dioxide are produced.

The power of stoichiometry isn't just in predicting amounts; it also enables chemists to scale reactions up or down and to adjust proportions to ensure the most efficient use of resources. It's fundamental for everything from cooking recipes to manufacturing drugs, highlighting why a deep understanding of stoichiometry is essential for anyone studying chemistry.