Chemical equation
From Academic Kids

de:Reaktionsgleichung es:Ecuación química fr:équation chimique
In chemistry, a chemical equation is a symbolic representation of a chemical reaction. For example, the combustion of methane in oxygen is
 CH_{4} + 2 O_{2} → CO_{2} + 2 H_{2}O,
and the reversible reaction of the Haber process is
 N_{2(g)} + 3H_{2(g)} ↔ 2NH_{3(g)} + ΔH.
Contents 
How Do You Read Chemical Equations?
The first number in a set such as 3H_{2}O tells how many molecules of the substance there is. The number in subtext following the element tells how many atoms of the element there are. The letters represent elements from the Periodic table.
ex. 3H_{2}O means 3 molecules, each containing 2 parts Hydrogen, 1 part Oxygen.
How Do You Balance Chemical Equations?
To balance chemical equations, think of both sides as a balance that can totter. When one side is heavier it falls and the other rises. The Chemical equation is the same thing: When the equation is balanced, neither the identities nor the mass of both the reactants and the products remain unchanged.
Generally, it is best to balance the most complicated molecule first. Most chemical equations can be balanced by inspection, that is, by trial and error.
Let's look at a few examples and walk through it:
Ex #1. Na + O_{2} → Na_{2}O
In order for this equation to be balanced, there must be 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. We solve this problem by putting a 2 in front of the Na on the left hand side, Like this:
2Na + O_{2} → Na_{2}O
So now we see, there are 2 Na's on the left and 2 Na's on the right. But what about the O's? We now must check to see if the O's are balanced on both sides of the equation. On the left hand side there are 2 O's and the Right hand side only has one. This is still an unbalanced equation. To fix this we must put a 2 in front of the Na_{2}O on the right hand side. Now our equation reads:
2Na + O_{2} → 2Na_{2}O
Notice that the 2 on the right hand side is "distributed" to both the Na_{2} and the O. Currently the left hand side of the equation has 2 Na's and 2O's. 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 let's add 2 more Na's on the left side. The equation will now look like this:
4Na + O_{2} → 2Na_{2}O
So now, as you can see, we have a balanced equation because there is an equal amount of element's on the left and right hand sides of the equation.
Ex #2. P_{4} + O_{2} → P_{4}O_{10}
This equation is not balanced becuse 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's balance. Let's look at the O's. The left hand side has 2 O's and the right hand side has 10 O's. To fix this unbalanced equation we must put a 5 in front of the O_{2} on the left hand side to make 10 O's on both sides. Let's take a look at what this looks like:
P_{4} + 5O_{2} → P_{4}O_{10}
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. C_{2}H_{5}OH + O_{2} → CO_{2} + H_{2}O
This equation is more complex than the previous examples; it will take a few steps. The most complicated molecule here is C_{2}H_{5}OH, so we begin by placing the coefficient 2 before the CO_{2} to balance the carbon atoms.
C_{2}H_{5}OH + O_{2} → 2CO_{2} + H_{2}O
Since C_{2}H_{5}OH contains 6 hydrogen atoms, the hydrogen atoms can be balanced by placing 3 before the H_{2}O, like this:
C_{2}H_{5}OH + O_{2} → 2CO_{2} + 3H_{2}O
Finally we balance the oxygen atoms. Since there are 7 oxygen atoms on the right and only 3 on the left, we balance it by placing a 3 before O_{2}, to produce the balanced equation:
C_{2}H_{5}OH + 3O_{2} → 2CO_{2} + 3H_{2}O
Weblinks
 Online calculator (http://sciencesoft.at/index.jsp?link=solve&lang=en), which deals with the determination of the coefficients of a chemical equation: