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A chemical equation is a symbolic representation of a chemical reaction. ^[1] The coefficients next to the symbols and formulae of entities are the absolute values of the stoichiometric numbers. The first-ever chemical equation was diagrammed by Jean Beguin in 1615.

Contents 1 Overview 2 Balancing chemical equations 2.1 Linear system balancing 3 Reading chemical equations 4 References 5 External links

Different symbols are used to connect the reactants and products with the following meanings: = for a stoichiometric relation; → for a net forward reaction; ⇆ for a reaction in both directions; ⇌ for equilibrium

For example, the combustion of methane (in oxygen) is depicted as:

CH₄ + 2 O₂ → CO₂ + 2 H₂O

In a chemical reaction, the quantity of each element does not change. Thus, each side of the equation must represent the same quantity of any particular element. Also in case of net ionic reactions the same charge must be present on both sides of the hiddly unbalanced equation, one may balance it by changing the scalar number for each molecular formula.

(What is a hiddly unbalanced equation?)

Simple chemical equations can be balanced by inspection, that is, by trial and error. Generally, it is best to balance the most complicated molecule first. Hydrogen and oxygen are usually balanced last.

Ex #1. Na + O₂ → Na₂O

In order for this equation to be balanced, there must be an equal amount of Na on the left hand side as on the right hand side. As it stands now, there is 1 Na on the left but 2 Na's on the right. This problem is solved by putting a 2 in front of the Na on the left hand side:

2Na + O₂ → Na₂O

In this there are 2 Na atoms on the left and 2 Na atoms on the right. In the next step the oxygen atoms are balanced as well. On the left hand side there are 2 O atoms and the right hand side only has one. This is still an unbalanced equation. To fix this a 2 is added in front of the Na₂O on the right hand side. Now the equation reads:

2Na + O₂ → 2Na₂O

Notice that the 2 on the right hand side is "distributed" to both the Na₂ and the O. Currently the left hand side of the equation has 2 Na atoms and 2 O atoms. The right hand side has 4 Na's total and 2 O's. Again, this is a problem, there must be an equal amount of each chemical on both sides. To fix this 2 more Na's are added on the left side. The equation will now look like this:

4Na + O₂ → 2Na₂O

This equation is a balanced equation because there is an equal number of atoms of each element on the left and right hand sides of the equation.

Ex #2. This equation is not balanced because there is an unequal amount of O's on both sides of the equation. The left hand side has 4 P's and the right hand side has 4 P's. So the P atoms are balanced. The left hand side has 2 O's and the right hand side has 10 O's.

P₄ + O₂ → P₄O₁₀

To fix this unbalanced equation a 5 in front of the O₂ on the left hand side is added to make 10 O's on both sides resulting in

P₄ + 5O₂ → P₄O₁₀

The equation is now balanced because there is an equal amount of substances on the left and the right hand side of the equation.

Ex #3. C₂H₅OH + O₂ → CO₂ + H₂O

This equation is more complex than the previous examples and requires more steps. The most complicated molecule here is C₂H₅OH, so balancing begins by placing the coefficient 2 before the CO₂ to balance the carbon atoms.

C₂H₅OH + O₂ → 2CO₂ + H₂O

Since C₂H₅OH contains 6 hydrogen atoms, the hydrogen atoms can be balanced by placing 3 before the H₂O:

C₂H₅OH + O₂ → 2CO₂ + 3H₂O

Finally the oxygen atoms must be balanced. Since there are 7 oxygen atoms on the right and only 3 on the left, a 3 is placed before O₂, to produce the balanced equation:

C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O

In reactions involving many compounds, balancing may get harder, we can then try to balance equation using algebraic method, based on solving set of linear equations:

1. Assign variables to each coefficient (coefficients represent both the basic unit and mole ratios in balanced equations):

• a K₄Fe(CN)₆ + b H₂SO₄ + c H₂O → d K₂SO₄ + e FeSO₄ + f (NH₄)₂SO₄ + g CO

2. We must have the same quantities of each atom in each side of the equation. So, for each element, count its atoms and equal both sides:

• K: 4a = 2d
• Fe: 1a = 1e
• C: a = 6g
• N: a = 3f
• H: 2b+2c = 8f
• S: b = d+e+f
• O: 4b+c = 4d+4e+4f+g

3. Solving the system (usually direct substitution is the best way)

• d=2a
• e=a
• g=6a
• f=3a
• b=6a
• c=6a

which means that we have all coefficients depending on a parameter a, just choose a=1 (a number that will make all of them small whole numbers) and you'll have:

• a=1 b=6 c=6 d=2 e=1 f=3 g=6

4. And the balanced equation at last:

• K₄Fe(CN)₆ + 6 H₂SO₄ + 6 H₂O → 2 K₂SO₄ + FeSO₄ + 3 (NH₄)₂SO₄ + 6 CO

To speed up the process, one can combine both methods to get a more practical algorithm:

1. Identify elements which occur in one compound in each member (this is very usual)

2. Start with the one among those which has a big index (this will help to keep working with integers), and assign a variable, let's say a.

• a K₄Fe(CN)₆ + H₂SO₄ + H₂O → K₂SO₄ + FeSO₄ + (NH₄)₂SO₄ + CO

3. Well, K₂SO₄ has to be 2a (because of K), and also, FeSO₄ has to be 1a (because of Fe), CO has to be 6a (because of C) and (NH₄)₂SO₄ has to be 3a (because of N). Well, this takes out the first four equations of the system! We already know that, whatever the coefficients are, those proportions must hold:

• a K₄Fe(CN)₆ + H₂SO₄ + H₂O → 2a K₂SO₄ + a FeSO₄ + 3a (NH₄)₂SO₄ + 6a CO

4. We can continue by writing the equations now (and having simpler problem to solve) or, in this particular case (although not so particular) we could continue by noticing that adding the Sulfurs we get 6a for H₂SO₄ and finally by adding the hydrogens (or the oxygens) we get the lasting 6a for H₂SO₄.

5. Again, having a convenient value for a (in this case 1 will do, but if a gets fractionary values in the other coefficients you will like to cancel the denominators) we get the result:

• K₄Fe(CN)₆ + 6 H₂SO₄ + 6 H₂O → 2 K₂SO₄ + FeSO₄ + 3 (NH₄)₂SO₄ + 6 CO

When reading a chemical equation there are some points to consider.

• Each side of an equation represents a mixture of chemicals. The mixture is written as a set of molecular formulas, separated by + symbols.
• Each formula is preceded by an optional scalar number (if no scalar number is written, it is implied that the number is 1). The scalar numbers indicate the relative quantity of molecules in the reaction. For instance, the string 2H₂O + 3CH₄ represents a mixture containing 2 molecules of H₂O for every 3 molecules of CH₄.
• The two sides of the equation are separated by an arrow. If the reaction is non-reversible, a right-arrow (→) is used, indicating that the left side represents the mixture of chemicals before the reaction, and the right side indicates the mixture after the reaction. For a reversible reaction, a two-way arrow is used. For example the equation 4Na + O₂ → 2Na₂O represents a non-reversible reaction. In this reaction, sodium (Na) and oxygen (O₂) are converted to a single molecule, Na₂O (containing 2 sodium atoms and 1 oxygen atom). We can also see that for every 4 sodium atoms at the beginning of the reaction, a single O₂ molecule will participate, and 2 Na₂O molecules will result.
• A chemical equation does not imply that all reactants are consumed in a chemical process. For instance a limiting reagent determines how far a reaction can go.
• In an ionic equation balancing of charge also takes place. In a full equation all reactants are written as molecules.

1. ^ IUPAC Compendium of Chemical Terminology

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