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### Balancing Equations

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

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

A. This is not a difficult task and can easily be accomplishedusing some basic problem solving skills. In fact, what follows is achemistry 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 andproducts in the reaction.

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

3. Count the number of atoms of each element in the reactants aproducts. A polyatomic ion appearing unchanged on both sides of theequation 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 frontof a formula a an equation. When no coefficient is written, it isassumed to be 1. It is best to begin with an element other thanhydrogen or oxygen. These two elements often occur more than twice inan equation.You must not attempt to balance an equation bychanging the subscripts in the chemical formula of a substance.5. Check each atom or polyatomic ion to be sure that the equationis balanced.

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

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

Example 3

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

Solution

Since the chemical formulas of the reactants and products areknown, 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 of2 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 thereare in the reactants. To correct this, put a coefficient of 2 infront 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 possibleratio 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. Itwill not work the same way every time, and, for more complexequations, it may prove very frustrating. Another way to balanceequations is to mathematically solve them as a linear system ofequations. This method is more time-consuming than the previousmethod, 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!!

**II. Keeping this in mind let's again balance H**

_{2}+ O_{2}---> H_{2}O: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 thecoefficient 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 thesmallest 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.

**III. Balance C**

_{2}H_{6}O + O_{2}-----> CO_{2}+ H_{2}OA. 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 bestsince 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 areinvolved).

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 solveunder 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 eveneasier.

**IV. Use the new math method to balance the followingequations:**

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}