**Chemical reactions **are represented by chemical equation, which identify reactants and products.

Formulas of the reactants appear on the left side of the equation; those of products are written on the right.

In many cases it is useful to indicate the state or phases of the substance in an equation.You use the following phase labels:

(g) = gas , (l) = liquid , (s) = solid , (aq) = water solution

For example; Burning of sodium in chlorine produces sodium chloride.

2Na(s) + Cl2 (g) ------- > 2NaCl (s)

**Balancing equation:**

In a balanced chemical equation , there are same number of atoms of a given element on both sides. For that reason , any calculation involving a reaction must be based on the balanced equation for that reaction.

For example; You have to write a balanced equation when liquid hydrazine and dinitrogen tetraoxide, produces gaseous nitrogen and water vapour.

To write a balanced equation proceed as follows:

**1 . Write a "skeleton" equation in which the formulas of the reactants appear on the left and those of the products on the right.** In the above case,

N2H4 + N204 ---------> N2 + H2O

**2 . Indicate the physical state of each reactant and product, after the formula, by writing**

(g) for gas, (l) for pure liquid , (s) for solid , (aq) for ion or molecule in water solution

N2H4(l) + N204 (l) ---------> N2 (g) + H2O (g)

**3. Balance equation.**To accomplish this , start by writing a coefficient of 4 for H2O , thus obtaining 4 oxygen atoms on both side:

N2H4(l) + N204 (l) ---------> N2 (g) + 4H2O (g)

Now consider the hydrogen atoms.There are 4x2 = 7 H atoms on the right.To obtain 8 H atoms on the left , write a coeffcient of 2 for N2H4:

2N2H4(l) + N204 (l) ---------> N2 (g) + 4H2O (g)

Finally, consider nitrogen.There are a total of (2x2) + 2 = 6 nitrogen atoms on the left.To balance nitrogen atoms, write a coefficent of 3 for N2 :

2N2H4(l) + N204 (l) ---------> 3 N2 (g) + 4H2O (g)

This is final balanced equation for the reaction of hydrazine with dinitrogen tetraoxide.

**Three points concerning the balancing process are worth nothing:**

**1**.Equations are balanced by adjusting coefficents in front of formulas, never by changing subscripts within formulas. On the paper, the equation discussed above could have been balanced by writing N6 on the right , but that would have been absurd . Elemental nitrogen exists as diatomic molecule N2 ;there is no such thing as an N6 molecule.

2. In balancing an equation, it is best to start with an element that appears in only one species on each side of the equation.In this case,either oxygen or hydrogen is a good starting point.Nitrogen would have been a poor choice, however because there are nitrogen atoms in both reactant molecules , N2H4 and N2O4.

3. In principal, an infinite number of balanced equations can be written for any reaction.The equations;

4N2H4(l) + 2N204 (l) ---------> 6 N2 (g) + 8H2O (g)

N2H4(l) + 1/2N204 (l) ---------> 3/2 N2 (g) + 2H2O (g)

are balanced in that there are the same number of atoms of each element on both sides.Ordinarily, the equation with the simplest whole number coefficients

2N2H4(l) + N204 (l) ---------> 3 N2 (g) + 4H2O (g)

is preferred.

For more balancing related problems see our tutorial balancing equations: solved problems