Balancing a chemical equation

Chemistry Tutorial

Balancing a chemical equation

Intro

A common skill required in chemistry is recognizing and balancing chemical equations. A chemical reaction is written out in a format that displays compounds and/or individual elements with reactants on the left side of the arrow and products to the right of the arrow.

Sample Problem

CH4 + O2 –> CO2 + H20

This is the bare-bone equation (reaction that is yet to be balanced). It represents methane with the addition of oxygen produces carbon dioxide and water (side note: all combustion reactions produce those two compounds). Numbers between and after letters represent the amount of the particular element that is present and would normally be in subscript format. Balancing an equation involves placing coefficients in front of compounds or individual elements to ensure the same number of a given substance (such as oxygen) appears on both the reactant and product sides of the equation.

Solution

CH4 + O2 –> CO2 + H20

As it appears initially, there are 2 molecules of Oxygen, 4 molecules of Hydrogen, and 1 molecule of Carbon on the reactant side. On the product side, there are 3 molecules of Oxygen, 2 molecules of Hydrogen and again 1 molecule of carbon. So, in order to alter the equation so that the same number of a given molecule appears on both sides, coefficients must be added to compounds. What this means is placing the appropriate number in front of the compound which acts as a multiplier for the proceeding number of molecules. For example, since there are 4 Hydrogens on the reactant side of the equation and only 2 on the product side, a coefficient of 2 is placed in front of water like so:

CH4 + O2 –> CO2 + 2H20

This results in 2 x H2 = 4 Hydrogens AND 2 x O = 2 Oxygens since the coefficient applies to both molecules within the compound. However, this further imbalances the number of Oxygen molecules on either side of the equation. There are now 4 on the product side (2 molecules of Oxygen within Carbon Dioxide + 2 molecules of Oxygen within Water). To fix this, a coefficient of 2 may be added in front of Oxygen on the reactant side:

CH4 + 2O2 –> CO2 + 2H2O

Now, there are four molecules of Oxygen, four molecules of Hydrogen, and 1 molecule of Carbon on either side of the equation. The equation is now balanced.



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