Imagine you mix 16.05 g of methane (CH4) gas and 96.00 g of oxygen (O2) gas and then ignite the mixture. After a bright flash and a loud bang, some water droplets form on the inside of the reaction vessel. a. Write the balanced chemical reaction for the combustion of methane. b. Sketch the process that occurred in the vessel using circles to represent atoms. Represent carbon with black circles, hydrogen with white circles, and oxygen with red 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? What? How much? e. Identify the following: limiting reactant, reactant in excess, and theoretical yield.

Step by step solution

01

Balancing the Chemical Equation

The chemical reaction for the combustion of methane (CH4) with oxygen (O2) is CH4 + O2 -> CO2 + H2O. To balance the equation: Write down the number of atoms of each element present in the reactants and products (C:1, H:4, O:2 for reactants, C:1, H:2, O:3 for products). Balance the hydrogens by adjusting the coefficient of H2O to 2, giving us CH4 + O2 -> CO2 + 2H2O. Now balance the oxygens by adjusting the coefficient of O2 to 2, resulting in CH4 + 2O2 -> CO2 + 2H2O.

02

Sketching the Molecular Process

Represent each atom as a circle with the specified colors: black for carbon, white for hydrogen, and red for oxygen. Methane should be drawn as one black circle (C) with four white circles (H) bonded to it. Oxygen should be drawn as two red circles (O) bonded together. Sketch reactants on the left side and products on the right side, using the balanced chemical equation as a guide. For methane and oxygen, draw the number of molecules proportional to the coefficients in the balanced reaction. Similarly, draw the products (carbon dioxide and water) molecules with the corresponding colored circles and bonds.

03

Calculating Moles of Water and Carbon Dioxide Produced

Use the molar masses to convert the mass of methane and oxygen to moles. Methane (CH4) has a molar mass of approximately 16.04 g/mol, and oxygen (O2) has a molar mass of approximately 32.00 g/mol. The number of moles of methane is 16.05 g / 16.04 g/mol = 1.0006 moles. The number of moles of oxygen is 96.00 g / 32.00 g/mol = 3 moles. According to the stoichiometry of the balanced equation, 1 mole of methane reacts with 2 moles of oxygen to produce 1 mole of CO2 and 2 moles of H2O. Therefore, we can produce 1.0006 moles of CO2 and 2 * 1.0006 moles of H2O.

04

Identifying Leftover Reactants

Since 2 moles of O2 react with each mole of CH4 and we have 3 moles of O2 available for 1.0006 moles of CH4, we have an excess of O2 after the reaction. 3 moles - 2 * 1.0006 moles = 1.0006 mole of O2 is left over.

05

Determining Limiting Reactant, Reactant in Excess, and Theoretical Yield

The limiting reactant is the reactant that will be completely consumed first during the chemical reaction, which is methane (CH4) in this case. The reactant in excess is oxygen (O2) because there is some left after the reaction is complete. The theoretical yield can be determined by the amount of products formed based on the limiting reactant. Therefore, the theoretical yield of water is 2 * 1.0006 moles of H2O, and for carbon dioxide, it is 1.0006 moles of CO2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Reaction Balancing

Chemical reaction balancing is a crucial step in understanding chemical equations. For a reaction to comply with the law of conservation of mass, there must be an equal number of each type of atom on both sides of the equation. When balancing the reaction for the combustion of methane, we take the unbalanced equation CH4 + O2 -> CO2 + H2O and strategically adjust the coefficients until the number of atoms of each element is the same on both sides.

This is not only critical for clear communication in chemistry but also sets the foundation for further calculations involving stoichiometry, determining the limiting reactant, and predicting theoretical yield in chemical processes.

Stoichiometry

Stoichiometry is like a recipe for chemistry; it tells us the exact proportions of reactants needed to produce a desired amount of product in a chemical reaction. In the stoichiometric analysis of the combustion of methane, we use the balanced chemical equation to figure out the relationship between the substances. For example, we determine that 1 mole of methane reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 moles of water.

With stoichiometry, we are able to translate between the mass of reactants and the moles of products, making it a vital tool for any chemist or student to predict how much product can be formed from given reactants.

Limiting Reactant

In a chemical reaction, the limiting reactant is the one that will be used up first, thus determining when the reaction stops and how much product is formed. To identify it, we need to consider the balanced chemical equation and the amount of each reactant available. In our methane combustion example, methane is the limiting reactant, as it will be completely consumed before all of the oxygen is used up.

Understanding the limiting reactant concept is critical because it helps us predict how much product can be formed and ensures efficient use of resources in chemical industries.

Reactant in Excess

When we talk about the reactant in excess, we are referring to the substance that remains after the limiting reactant has been fully consumed in a chemical reaction. There is more of the reactant in excess than is necessary to completely react with the limiting reactant. In this methane combustion scenario, oxygen is in excess; after all the methane has reacted, there is still some oxygen left over.

This knowledge can be particularly useful when considering the practical implications of a reaction, such as the safety concerns of storing reactive leftovers or the need to purify the products.

Theoretical Yield

The theoretical yield is the maximum amount of product that can be generated from a given amount of reactant, assuming the reaction goes to completion as described by the stoichiometry. It represents an ideal scenario, with no loss of material and no side reactions. When calculating the theoretical yield of the combustion of methane, we base it on the amount of methane present, since it is our limiting reactant.

The theoretical yield is essential for planning reactions and evaluating the efficiency of experimental procedures in both educational settings and industry. For our methane combustion reaction, it tells us exactly how much carbon dioxide and water we would expect to see in a perfect world.