Download Balancing Chemical Equations

Balancing Chemical Equations
A mathematical equation is simply a sentence that states that two expressions are equal.
One or both of the expressions will contain a variable whose value must be determined by
solving the equation. Linear equations are equations in which the variable being solved for
is raised to a power no higher than 1. Equations in which the variable being solved for is
raised to a power of 2 are called quadratic equations. The problems in this text do not
require you to solve quadratic equations, so this topic will not be covered here. Consider
the following linear equation.
5x - 3 = 2x + 9
To solve the equation for x, the first step is to get all of the terms involving x on one side of
the equation, and all other terms on the other side. We are allowed to perform the
following operations on any equation without changing the equality that the equation
represents.
1. Adding the same quantity to both sides of the equation.
2. Subtracting the same quantity from both sides of the equation.
3. Multiplying both sides of the equation by the same quantity.
4. Dividing both sides of the equation by the same quantity.
The basic rule is: Whatever you do to one side, do to the other side also.
In the equation 5x - 3 = 2x + 9, we will collect all of the x’s on the left side of the
equation and all of the numbers on the right side. We must add 3 to each side to get the
numbers on the right side, and subtract 2x from each side to get the x’ on the left side.
Original equation:
(1) add three to both side:
5x - 3 = 2x + 9
+3
+ 3
5x
= 2x + 12
-2x
-2x
3x
=
12
(2) subtract 2x from both sides:
Now to solve for x, we must divide both sides of the equation by 3.
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(3) divide both sides by 3:
3x = 12
3
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Final Answer:
x
= 4
Balancing Equations Example 1
Solve for the value of x in the equation below.
6x + 7 = 4x + 23
Solution:
First, subtract 7 from both sides. Then, subtract 4x from both sides. Finally, divide
both sides by 2 to get the final answer: x = 8. You will be on track to get the right
answer if you remember one simple rule: Whatever you do to one side, do to the
other side also.
Now what does this have to do with balancing equations? You can use basic algebra
rearrangements like we did in the example above to help you balance equations. Look at
this equation.
Ni(s) + HCl(aq) NiCl2(aq) + H2(g)
A balanced equation must have the same number of each kind of atom on both sides of
the equation. The same number of nickel atoms must appear on both sides of the
equation. The same number of hydrogen atoms must appear on both sides of the
equation. The same number of chlorine atoms must appear on both sides of the equation.
As it is currently written, the equation is not balanced. The nickel atoms are balanced, but
the hydrogen and chlorine atoms are not balanced. Let’s balance the chlorine atoms using
some algebra.
There is one chlorine atom on the left side of the equation and two chlorine atoms on the
right side of the equation. What do you need to multiply the left side of the equation by in
order to make the two sides of the equation have the same number of chlorine atoms? Let
this unknown multiplication factor be “x”. We can set up an algebraic equation for this
question that looks like:
x(1) = 2
2
To solve for x, divide both sides by one. The final answer is x = 2. If we put a coefficient
of two in front of HCl, the equation becomes:
Ni(s) + 2HCl(aq) NiCl2(aq) + H2(g)
The equation is now balanced.
That was a very simple application of algebra to the problem of balancing an equation.
Let’s look at a more complex example.
Fe(s) + O2(g) Fe2O3(s)
Let’s balanced the iron atoms first. There is one iron atom on the left side of the equation
and two iron atoms on the right side of the equation. The algebra expression that
represents this situation would be:
x(1) = 2
Solving for x by dividing both sides by one gives us the final answer of: x = 2. We need to
put a coefficient of two in front of Fe(s) on the left to balance the iron atoms.
2 Fe(s) + O2(g) Fe2O3(s)
There are two oxygen atoms on the left side of the equation and three oxygen atoms on
the right side of the equation. The algebra expression that represents this situation would
need to say that some unknown multiplication factor, y, times the two oxygen atoms on the
left will equal the three oxygen atoms on the right.
y(2) = 3
Solve for y by dividing both sides by two. The final answer is y = 3/2. We need to put a
coefficient of 3/2 in front of the O2 on the left to balance the oxygen atoms. It is easier to
leave this answer in the fractional form rather than convert it to a decimal number. You’ll
see why soon.
2 Fe(s) + 3/2 O2(g) Fe2O3(s)
The equation is now balanced. However, as often happens, a fraction appears in the
balanced equation. Chemical reactions, except in a few limited circumstances, should not
be balanced with fractions. In order to eliminate the fraction, we need to multiply all of the
coefficients by the number in the denominator. Remember the basic rule: Whatever you
do to one side, do to the other side also.
.
.
2 2Fe(s) + 2 3/2 O2(g) 2 Fe2O3(s)
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The equation becomes:
4 Fe(s) + 3 O2(g) 2 Fe2O3(s)
Balancing Equations Example 2
Balance this equation:
HF(g) + SiO2(s) SiF4(g) + H2O(l)
Solution:
If chemical equation is difficult to balance, you can use simple algebra expressions
to make balancing the equation easier. To balance the fluorine atoms, you could
use the algebra expression:
x(1) = 4
Solving for x gives you a final answer of: x = 4. You need a coefficient of four in
front of HF(g). To balance the oxygens, you could use the algebra expression:
2 = y(1)
Solving for y gives you a final answer of: 2 = y. You need a coefficient of two in
front of H2O(l).
The balanced equation is:
4HF(g) + SiO2(s) SiF4(g) + 2H2O(l)
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Balancing Equations Example 3
Balance the equation
C2H6(g) + O2(g) CO2(g) + H2O(l)
Solution:
Begin by balancing the carbons. The algebra expression for carbons could be 2 =
x(1). Solving for x gives the final answer: 2 = x. You need a coefficient of two in
front of CO2(g). The algebra expression for hydrogen could be 6 = y(2). Solving for
y gives the final answer: 3 = y. You need a coefficient of 3 in front of H2O(l).
C2H6(g) + O2(g) 2 CO2(g) + 3 H2O(l)
The algebra expression for oxygen is more complex. There are two oxygen atoms
on the left side of the equation and 4 + 3 oxygen atoms on the right side of the
equation. The algebra expression could be:
z(2) = 7
Solving for z gives the final answer: z = 7/2. The balanced equation is:
C2H6(g) + 7/2 O2(g) 2 CO2(g) + 3 H2O(l)
Although the equation is balanced, we need to eliminate the fraction from the
chemical equation. We do this by multiplying coefficients on both sides of the
equation by whatever value is in the denominator. In this case, we would multiply
by two.
.
.
.
2 C2H6(g) + 2 7/2 O2(g) 2 2 CO2(g) + 2 3 H2O(l)
The equation becomes
2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(l)
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Hints for Balancing Equations:
Do I need to use algebra to balance an equation?
No. Most students quickly learn how to balance the simpler chemical equations.
However, if a equation is difficult or tricky to balance, writing out algebra expression for
each different kind of atom is a reliable way to balance an equation.
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