#### BALANCING EQUATIONS

The chemical equation is a short-hand method for recording which chemical substances reacted during a chemical reaction and which substances are produced. Exact chemical formulas are used to represent the substances that are involved in the reaction equation.

The chemical substances that are reacted (added) together are listed on the left-hand side of the equation and are separated by the plus (+) sign. The chemical substances that are produced as a result of the reaction are noted on the right-hand side of the equation and each substance is separated by a plus sign. An arrow ( ) is generally used to indicate that the reaction proceeds from the reactants to the products.

The balanced chemical equation is like the recipe for chemical reaction. It provides information as to the exact amounts required of each reactant to react with another reactant in order for the reaction to be completed.

Exact chemical formulas are formulas that are derived by utilizing the oxidation numbers (charges) of each ion to derive the formulas for that compound.

(1) H_{2} + Cl_{2} ? HCl

A balanced equation requires that the number of atoms on the left-hand side of the equation equal that number of atoms on the right-hand side of the equation.

In equation (1) There are two atoms of H on the left, there must be two atoms on the right.

The coefficient is used to balance the chemical equations. The subscripts contained in the formulas of a compound should never be changed. They are derived when the formulas are calculated. You can put a “**2**” in front of the HCl formula. This indicates that you have two formulas of HCl; two H’s as well as two Cl’s.

H_{2} + Cl_{2} ? ** 2** HCl

This equation is balanced because. There are two atoms of H on the left; there are two atoms of H on the right. There are two atoms of Cl on the left; there are two atoms of Cl on the right.

(2) H_{2}SO_{4} + NaOH ? Na_{2}SO_{4} + H_{2}O

You should first look to see which atoms are not equal on both sides of the equation.

There are three “H” on the left, two on the right

There is one “SO_{4}” on the left, one on the right. It is convenient to treat radicals as a whole rather than as separate atoms.

Not counting the “O” in SO4, there is one other “O” on the left, one on the right

This is one “Na” on the left, two on the right

Initially, the Na’s and “O” are not balanced. You will have to use coefficients and only coefficients to help balance the equations. Many times changing the coefficient on one side causes an atom that was previously balanced to be unbalanced. However, you must toggle back and forth until you have balanced all of the atoms. Let’s begin….

Choose which atom you would like to balance first, many times the other will be balanced

We will first put a “**2**” as a coefficient for NaOH to balance the Na’s

H_{2}SO_{4} + **2** NaOH ? Na_{2}SO_{4} + H_{2}O

Now, the Na’s are balanced, but the H’s have become unbalanced. There are four on the left, only two on the right.

We now will put a coefficient of “**2**” in front of the H_{2}O.

H_{2}SO_{4} + **2** NaOH ? Na_{2}SO_{4} + **2**H_{2}O

The equation is now balanced.

There are four “H” on the left, four on the right

There is one “SO_{4}” on the left, one on the right. It is convenient to treat radicals as a whole rather than as separate atoms.

Not counting the “O” in SO4, there is two other “O” on the left, two on the right

There are two “Na's” on the left, two on the right

(3) K + Cl_{2 + }O_{2} ? KClO_{3}

In this equation, initially the “Cl” and the “O” are not balanced. They do not have the same number of atoms on both sides of the equation.

There is one “K”, on the left, one on the right

There are two “Cl” on the left, one on the right

There are two “O” on the left, three on the right

When the number of atoms of a particular element on one side of the equation can not be balanced on the opposite side of the equation by applying a whole-number coefficient to that side, you will be required to use coefficients for each side of the equation to balance that element that are multiplicands of a least common product. For example, in the case above, there are three “O”’s on the right side, two on the left. You cannot find a whole number coefficient to multiply the left side to balance the “O”’s . The first trial should be to multiply 2 x 3 to get a product. You use the least common product such as 6 to determine what to multiply each side. The side with two “O” ‘s will be multiplied by three; the side with three “O”’s will be multiplied by two.

K + Cl_{2 + }**3**_{ }O_{2} ? **2** KClO_{3}

The “O”’s and “Cl”’s are balanced, now the K’s must be balanced next... We will use a two as the coefficient for K on the left side.

**2** K + Cl_{2 + }**3**_{ }O_{2} ? **2** KClO_{3}