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Balancing chemical equations

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.

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

[edit] Linear system balancing

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


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[Golden_Rhino]

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[Paulie Walnuts]

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