Balancing equations is a fundamental skill in Chemistry. Solving a system of linear equations is a fundamental skill in Algebra. Remarkably, these two field specialties are intrinsically and inherently linked.

**I. Balance H _{2} + O_{2} ---->
H_{2}O**

A. This is not a difficult task and can easily be accomplished using some basic problem solving skills. In fact, what follows is a chemistry text's explanation of the situation:

Rules for Balancing EquationsTaken from:

Chemistry

Wilberham, Staley, Simpson, Matta

Addison Wesley1. Determine the correct formulas for all the reactants and products in the reaction.

2. Write the formulas for the reactants on the left and the formulas for the products on the right with an arrow in between. If two or more reactants or products are involved, separate their formulas with plus signs.

3. Count the number of atoms of each element in the reactants a products. A polyatomic ion appearing unchanged on both sides of the equation is counted as a single unit.

4. Balance the elements one at a time by using coefficients. A

coefficientis a small whole number that appears in front of a formula a an equation. When no coefficient is written, it is assumed to be 1. It is best to begin with an element other than hydrogen or oxygen. These two elements often occur more than twice in an equation.You must not attempt to balance an equation by changing the subscripts in the chemical formula of a substance.5. Check each atom or polyatomic ion to be sure that the equation is balanced.

6. Finally, make sure that all the coefficients are in the lowest possible ratio.

Now let's use these rules to balance the equation for the formation of water from hydrogen and oxygen.

Example 3

When hydrogen and oxygen react, the product is water. Write a balanced equation for this reaction.

Solution

Since the chemical formulas of the reactants and products are known, we can write a skeleton equation.

H

_{2}(g) + O_{2}(g) H_{2}O(l)Hydrogen is balanced but oxygen is not. If we put a coefficient of 2 in front of H

_{2}O, the oxygen becomes balanced.H

_{2}(g) + O_{2}(g) 2H_{2}O(l)Now there are twice as many hydrogen atoms in the product as there are in the reactants. To correct this, put a coefficient of 2 in front of H

_{2}. The equation is now balanced.2H

_{2}(g) + O_{2}(g) 2H_{2}O(l)Check the coefficients. They must be in their lowest possible ratio 2(H

_{2})s 1(O2), and 2(H_{2}O).End of Quotation

B. Not a bad process. However, it is not an exact process. It will not work the same way every time, and, for more complex equations, it may prove very frustrating. Another way to balance equations is to mathematically solve them as a linear system of equations. This method is more time-consuming than the previous method, but, it has several advantages:

- It is more concrete.
- It is especially valuable in solving more complex equations.
- It is a method that can be easily programmed into a computer or calculator.
- It demonstrates a terrific connection between math and science!!

A. First, place letter coefficients in front of each term:
aH_{2} + bO_{2} ---> cH_{2}O

B. Now, use each element to produce an equation involving the coefficient letters:

Hydrogen: 2a + 0b = 2c

Oxygen: 0a + 2b = 1c

C. Next, use math to solve this system of equations:

2a=2c implies a=c

2b=c implies b=1/2 c

D. Now, choose any integer value for c (since we would like the smallest integral solutions, c=2 works well).

E. If c=2; then a=2 also; and b=1/2 c =(1/2)(2) =1

F. Thus, as before 2H_{2} + 1O_{2} ---->
2H_{2}O.

A. Coefficients: aC_{2}H_{6}O + bO_{2}
-----> cCO_{2} + dH_{2}O

B. Equations:

C: 2a + 0b = 1c + 0d

H: 6a + 0b = 0c + 2d

O: 1a + 2b = 2c + 1d

C. Solve in terms of any letter. In this case, "a" may work best since it is represented in each equation:

c=2a

2d=6a implies d=3a

2c+d-2b=a

by substituion: 2(2a) + 3a -2b =a implies 7a -2b =a

which implies: -2b = -6a implies b=3a

Therefore: b=3a; c=2a; and d=3a

D. Now let "a" be any integer (1 works well when no fractions are involved).

Thus, a=1; b=3; c=2; and d=3 and consequently

E. 1C_{2}H_{6}O + 3 O_{2} ----->
2CO_{2} + 3H_{2}O

F. This equation may have been difficult for a student to solve under the old method.

"Where do I start" might have been a common reply to the problem. Mastery of this mathematical method should elimnate that dilemma. Later, we will discover how matrices will make this process even easier.

1. Zn + HCl ----> ZnCl_{2} + H_{2}

2. Al + O_{2} -----> Al_{2}O_{3}

3. Al + CuSO_{4} ------>
Al_{2}(SO_{4})_{3} + Cu

4. Li + H_{2}O ------> LiOH + H_{2}