Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

What volume of oxygen is needed to completely combust 2.36 L of methane gas (CH4)?

Short Answer

Expert verified
The volume of oxygen needed to completely combust 2.36 L of methane gas is 4.72L.

Step by step solution

01

Write the balanced chemical equation for methane combustion

Start by writing the balanced chemical equation for the combustion of methane. Methane reacts with oxygen to produce carbon dioxide and water: CH4+2O2CO2+2H2O The balanced equation tells us that one mole of methane reacts with two moles of oxygen to produce one mole of carbon dioxide and two moles of water.
02

Use the stoichiometric coefficients to find the ratio of methane to oxygen

According to the balanced chemical equation, one mole of methane requires two moles of oxygen for complete combustion. This means that the volume of oxygen needed will be twice the volume of methane.
03

Calculate the volume of oxygen needed

From the stoichiometry of the reaction, we know that the volume of oxygen needed is twice the volume of methane. Given that the volume of methane is 2.36 L, we can calculate the volume of oxygen needed: Volume of oxygen = 2 * Volume of methane Volume of oxygen = 2 * 2.36 L Volume of oxygen = 4.72L Therefore, the volume of oxygen needed to completely combust 2.36 L of methane gas is 4.72L.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Chemical Equation Balancing
Balancing a chemical equation is a fundamental skill in chemistry that involves ensuring that the number of atoms of each element is the same on both sides of the equation. This is essential because the Law of Conservation of Mass dictates that matter cannot be created or destroyed in a chemical reaction.

For example, in the combustion of methane, the balanced chemical equation is CH4+2O2CO2+2H2O.This indicates that one molecule of methane (CH4) reacts with two molecules of oxygen (O2) to produce one molecule of carbon dioxide (CO2) and two molecules of water (H2O). Balancing equations requires practice, and one helpful tip is to start by balancing elements that appear only in one reactant and one product first, then move on to the more complex cases.
Combustion Reaction
A combustion reaction involves a substance (often a hydrocarbon) reacting with oxygen to release energy in the form of heat and light. Combustion reactions are exothermic, meaning they give off heat. They're also a major source of energy for various applications, such as in car engines and power plants.

In the case of methane, a hydrocarbon, the reaction releases carbon dioxide and water as byproducts, which is represented by the chemical equation CH4+2O2CO2+2H2O.Knowing the balanced equation for a combustion reaction allows us to understand the proportions of reactants needed and the amount of products formed, crucial for industrial processes and environmental considerations.
Mole-to-Mole Ratio
The mole-to-mole ratio derived from a balanced chemical equation provides the relationship between the amounts in moles of any two substances involved in a chemical reaction. This concept is a cornerstone of stoichiometry, which is the quantitative aspect of chemical reactions.

In the previously mentioned methane combustion reaction, the balanced chemical equation indicates a mole ratio of 1:2 between methane and oxygen, meaning 1 mole of methane reacts with 2 moles of oxygen. Therefore, for every liter of methane, we'd need two liters of oxygen since the molar volume of gases under the same conditions of temperature and pressure is constant. Applying this knowledge to a practical situation: if you have 2.36 liters of methane CH4, you'd therefore require 2×2.36=4.72 liters of oxygen for complete combustion, illustrating the direct application of mole-to-mole ratios in calculating reactant and product volumes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.

Sign-up for free