Home
Math by Grades Pre-K
Kindergarten
Grade 1
Grade 2
Grade 3
Grade 4
Grade 5
Grade 6
Grades 7 and 8
Grades 9 and 10
Grades 11 and 12
Math by Topics Arithmetic
Algebra
Geometry
Math Word Problems
Trigonometry
Statistics
Probability
PreCalculus
Calculus
Set Theory
Matrices
Vectors
Math Worksheets Math Worksheets
Interactive Zone
Math in Video Lessons Basic Algebra
Intermediate Algebra
College Algebra
High School Geometry
College Calculus
Linear Algebra
Engineering Math
Singapore Math
Math for Specific Tests SAT Math
ACT Math
GMAT Math
High School, Regents
California Standards
GCSE Maths
A Level Maths
Math Fun and Games Math Trivia
Math Games
Fun Games
Mousehunt Guide
Exam Preparation SAT Preparation
ACT Preparation
GMAT Preparation
Science Biology
Chemistry
Science Projects
High School Biology
High School Chemistry
High School Physics
GCSE Biology
Others English Help
ESL, IELTS, TOEFL
Programming
Animal Facts
Tutoring Services
What's New

 

Balancing chemical equations

Atoms can neither be destroyed nor created during a simple chemical reaction. Therefore, in a chemical reaction

The sum of atoms before reaction = the sum of atoms after reaction

 

 

General Rules for balancing chemical equations – Rule 4

Balancing chemical equations may require some trial and error. There are some general rules that could be helpful, but they may not work all the time.

 

 

In this page, we will look at some examples of applying

Rule 4: Balancing chemical equations using the even technique.

If you have an even number of a certain element on one side of the equation and an odd number of the same element on the other side of the equation, multiply both sides of the equation through by the coefficient of 2. This will give an even number on both sides and make the equation easier to balance.

 

 

Example:

Balance the equation

CH3OH + O2 → H2O + CO2

Solution:

Step 1: Using the CHO technique, we start with carbon, one on each side, so carbon is balanced. There are four H on the left and two on the right, so we place the coefficient of 2 in front of the H2O on the right

CH3OH + O2 → 2H2O + CO2

Step 2: When we try to balance the oxygen, we find three on the left and four on the right. We multiply both sides of the equation through by two.

2CH3OH + 2O2 → 4H2O + 2CO2

Step 3: The C and H are still balanced, and now there are six O on the left and eight on the right. Change the coefficient in front of the O2 to 3 to give eight O on the left.

2CH3OH + 3O2 → 4H2O + 2CO2

Step 4: Check that all the atoms balance and make sure that all coefficients are in the lowest-possible ratio.

 

 

Videos

Balancing Equations Even Odd #1


Balancing Equations Even Odd #2

 

 

 

Custom Search

 

We welcome your feedback, comments and questions about this site - please submit your feedback via our Feedback page.

 

© Copyright 2005, 2008 - onlinemathlearning.com
Embedded content, if any, are copyrights of their respective owners.

 

 

Custom Search