In the notes on Chemical Nomenclature, you learned how to write
chemical formulas for numerous compounds. In the notes on Counting
Atoms in Chemical Formulas, you learned how to determine the number of
atoms of each element in these formulas. It is now time to put our
knowledge of chemical formulas to use in chemical equations.

Chemistry is primarily concerned with the changes matter undergoes.
Some of these changes are merely physical, such as when ice melts to form
water. But the most interesting changes in chemistry are those in
which new substances are formed from "original" substances. These
are called **chemical reactions** or **chemical changes**.
In many chemical reactions, the new substances that are formed have properties
that are quite different from those of the substances we started with.
But regardless of how dramatic the reaction, or how different the properties
of the new substances are from those of the original substances, we know
that at the atomic level, a chemical reaction is just a rearrangement of
atoms. Just as the same letters of the alphabet can be be rearranged
to form two very different words, so too can the same atoms often by put
together in different ways to form substances with very different properties.

We write a chemical equation for a reaction by writing the substances
we start with, separating them by plus signs (+) if there is more than
one, using an arrow to represent the process of change, and then writing
the substances we end up with, again separating them with plus signs.
In order to write a chemical equation, we must know three things:

- what substances we start with
- what substances we end up with
- the chemical formulas of the substances in items 1 and 2 above

Let's now look at some examples of writing and balancing chemical equations.

**Problem:** Hydrogen (H_{2}) burns in oxygen (O_{2})
to form water (H_{2}O). Write a balanced chemical equation
for this reaction, showing all substances as gases.

**Solution:**

Lets begin by writing a word description of the problem. From the information given above, we have

hydrogen + oxygen ----------> water

I guess the above is not absolutely necessary, but if you are a beginner, it may help you to see your way. The next step is to replace each name with its corresponding chemical formula. The formulas were all given in the problem, so we have

H_{2}(g) + O_{2}(g)
----------> H_{2}O(g)

The letter g in parentheses after each formula in the equation above indicates that the substance is a gas. Other phase labels we will encounter are l for liquid, s for solid and aq for aqueous. The term aqueous means dissolved in water.

Looking at the equation we have so far, we can see it is not balanced. Keep in mind that a chemical reaction is just a rearrangement of atoms, so every atom you start out with has to be somewhere in the end. And you can't end up with any atoms you didn't have in the beginning. The equation above does not meet these requirements. Using the information in the notes on Counting Atoms in Chemical Formulas, we can see that the left side of this equation has 2 atoms of hydrogen and 2 atoms of oxygen, while the right hand side has 2 atoms of hydrogen and only 1 atom of oxygen. At this point, hydrogen is ok, but oxygen is not. In its present form, the equation suggests that an atom of oxygen has magically disappearred. A pictorial representation of the present form of this equation is shown in line A of Figure 1, where H atoms are red dots and O atoms are blue dots. Notice that we start with 2 blue dots and end up with only 1. This of course, is impossible.

Beginners are tempted to solve problems such as the one we now face
by adjusting the subscripts, but that is NEVER the thing to do! For
example, one incorrect approach someone might try is to drop the subscript
of 2 in O_{2} and write

H_{2}(g) + O(g) ---------->
H_{2}O(g)

The pictorial representation of this is shown in line B of Figure 1.
This DOES balance the equation, but that's not our only objective.
We also want our equation to be a correct description of what really happens
in nature. Oxygen does not exist as a monatomic gas. It is
known that the oxygen atoms are paried up in groups of 2 in naturally occurring
oxygen. Therefore, to have a correct description of the chemical
system, we MUST represent oxygen as O_{2} and not as O.

Another approach -- also incorrect -- that someone might try -- is to balance oxygen by adding a subscript 2 to the oxygen in water and write the equation as

H_{2}(g) + O_{2}(g)
----------> H_{2}O_{2}(g)

The pictorial representation of this is shown in line C of Figure 1.
Again, we have a balanced equation, but things are still not right.
Our original objective was to describe the formation of WATER, and H_{2}O_{2}
is **not** water. Often, when you start changing subscripts, you
get some non-sense formula that does not exist. In this case we do
get a real substance. H_{2}O_{2 }is known as hydrogen
peroxide. You might have a 3% solution of it in your medicine cabinet.
It's that aniseptic in the dark brown bottle. Still, we have not
achieved the original objective with this equation, so it's not the one
we want, even though it shows the formation of a real substance.

The bottom line is NEVER ADJUST SUBSCRIPTS! Every chemical substance has a definite, unchangable chemical formula. When you change subscrips, what you are doing is changing the chemical formula, which is never the correct course of action.

Rather than adjusting the subscripts, you can adjust the numbers that go in front of the formulas. These numbers are called coefficients, and they tell us how many of those formula units we have. The correctly balanced equation for the formation of water from hydrogen and oxygen is

2H_{2}(g) + O_{2}(g)
----------> 2H_{2}O(g)

The pictorial representation of this is seen in line D of Figure 1. It is the only version of the equation discussed so far that is correct. Notice that it is balanced. It has 4 atoms of hydrogen on both sides, and it has 2 atoms of oxygen on both sides. You should strive to learn how to obtain these atom counts by using the information you learned in the notes on Counting Atoms in Chemical Formulas. Since you have had limited exposure to these concepts at this point, however, I have included Figure 1 to help you. You can count the red (H) and blue (O) dots in line D of Figure 1 to confirm the atom counts I have given you. In addition to being balanced, it also achieves our objective of describing the reaction of hydrogen and oxygen -- as they naturally exist -- to produce water.

The balanced equation shows that the H_{2} and O_{2}
molecules can not react in a one-to-one ratio to produce water. Rather,
it takes twice as many molecules of H_{2} as you have molecules
of O_{2} to end up with only H_{2}O molecules and have
no H_{2} or O_{2 }molecules left over. We normally
balance equations using the smallest integers possible. This is called
balancing in lowest terms. Once you have a balanced equation, if
you multiply it throughout by any integer, you will still have a balanced
equation. For example, if you multiply our balanced equation

2H_{2}(g) + O_{2}(g)
----------> 2H_{2}O(g)

by 2, you get

4H_{2}(g) + 2O_{2}(g)
----------> 4H_{2}O(g)

This is still balanced. There are now 8 H atoms on each side and 4 O atoms on each side. Practice your skills in counting atoms in chemical formulas with this equation. If you can't reproduce the 8 and 4 I get in my atom count, then prove it to yourself by counting the red (H) and blue (O) dots in Figure 2.

Although I did present the balanced equation

2H_{2}(g) + O_{2}(g)
----------> 2H_{2}O(g)

some may still be wondering where I got it. Admittedly, it may seem that I sort of "pulled it out of a hat". The previous discussion was more focused on what is and is not a correct way to balance an equation, rather than presenting the thought process that one goes through step by step in actually balancing the equation. So I am going to balance the equation we have been working with again -- this time, not leaving out any steps.

Our starting equation is

H_{2}(g) + O_{2}(g)
----------> H_{2}O(g)

Let us begin with the hydrogen. We see that there are 2 hydrogens
on the left and two on the right. So far, so good. H atoms
are balanced. Then we look at oxygen. We have 2 O atoms on
the left, but only one on the right. Problem. We know that
if we are going to have 2 O atoms on the left, we must have 2 O atoms on
the right also. We are not allowed to add a subscript to H_{2}O,
because that would be changing the formula, so the only solution is to
form 2 molecules of H_{2}O instead of only 1. Our equation
now becomes

H_{2}(g) + O_{2}(g)
----------> 2H_{2}O(g)

O atoms are now balanced -- there are 2 O atoms on each side.
But now H atom are messed up. There are 2 H atoms on the left, but
4 H atoms on the right. If we are going to end up with 4 H atoms,
then we must start out with 4 H atoms. The way to acheive this is
to use 2 molecules of H_{2} instead of only 1. Now we have

2H_{2}(g) + O_{2}(g)
----------> 2H_{2}O(g)

At last, the equation is balanced. It has 4 atoms of H on both sides and 2 atoms of O on both sides. Your work is not done until ALL atoms are balanced, so be sure to check the status of each element after each balancing operation you do. Note that when an element does not balance, you try to increase the number of atoms of that element on the side with fewer atoms to make it equal the number of atoms of that element on the side with more atoms.

Balancing equations is somewhat of a trial and error method. It is not easy to give a "recipe" that will work in all situations. However, there are some general hints that can make the work easier. First balance those elements that appear in only one formula on each side of the equation. If a free element appears, try to balance those atoms last.

The equation forming water that we just balanced does not fit those rules too well, because both elements appear in just one formula on each side of the equation, and because both elements appear as free elements on the left hand side. I will now work several examples -- step by step -- that hopefully, well make the process more clear.

**Example 1**

Methane (CH_{4}) burns in oxygen (O_{2}) to form carbon
dioxide (CO_{2}) and water (H_{2}O). Write a balanced
equation for this reaction, showing all substances as gases.

**Solution:**

The word equation for this reaction is

methane + oxygen ----------> carbon dioxide + water

Replacing each name in the above equation with its formula gives

CH_{4}(g) + O_{2}(g)
----------> CO_{2}(g) + H_{2}O(g)

Following the hints I gave above, we would not want to start our attempts at balancing this by considering oxygen. It appears in more than one formula on the right hand side, so it would not be clear which coefficient to addjust. This leaves carbon and hydrogen as possibilities. I generally start with carbon, where C, H and O are involved.

Notice that there is 1 C atom on the left and 1 C atom on the right.
For now, at least, C is balanced. Now let's look at H atoms.
We have 4 H atoms in CH_{4 }but only 2 H atoms in H_{2}O.
We can get 4 H atoms on the right by forming two H_{2}O molecules
instead of only one. Our equation now becomes

CH_{4}(g) + O_{2}(g)
----------> CO_{2}(g) + 2H_{2}O(g)

H atoms are now balanced. There are 4 of them on each side.
C atoms are still balanced (one on each side). We must now look at
oxygen. We have 2 O atoms on the left in O_{2} and 4 O atoms
on the right -- two of them in CO_{2} and two of them in H_{2}O
(since we are producing 2 of these molecules). We can get 4 O atoms
on the left side of the equation by taking two O_{2} molecules
instead of only one. We now have

CH_{4}(g) + 2O_{2}(g)
----------> CO_{2}(g) + 2H_{2}O(g)

This equation is balanced. The number of atoms of each element is as follows: 1 C atom on each side, 4 H atoms on each side, and 4 O atoms on each side. Always remember to do a final check before considering your work complete. ALL atoms must balance.

**Example 2**

Propane (C_{3}H_{8}) burns in oyxgen (O_{2})
to form carbon dioxide (CO_{2}) and water (H_{2}O).
Write a balanced equation for this reaction, showing all substances as
gases.

**Solution:**

The word equation for the reaction is

propane + oxygen ----------> carbon dioxide + water

Replacing each chemical name with its formula (and showing it as a gas) gives

C_{3}H_{8}(g) + O_{2}(g)
----------> CO_{2}(g) + H_{2}O(g)

Starting with carbon, we see that there are 3 C atoms on the left but
only 1 C atom on the right. We put a coefficient of 3 in front of
CO_{2} to balance carbon. Our equation now becomes

C_{3}H_{8}(g) + O_{2}(g)
----------> 3CO_{2}(g) + H_{2}O(g)

Carbon is now balanced, with 3 C atoms on each side. We now look
at hydrogen. There are 8 Hatoms on the left (in C_{3}H_{8})
but only 2 H atoms on the right (in H_{2}O). We can get 8
H atoms on the right by putting a coefficient of 4 in front of H_{2}O.
Our equation now becomes

C_{3}H_{8}(g) + O_{2}(g)
----------> 3CO_{2}(g) + 4H_{2}O(g)

The final step is to balance oxygen. We have 2 O atoms on the
left (in O_{2}) and 10 O atoms on the right (6 of them are contained
in CO_{2} molecules and 4 of them are contained in H_{2}O
molecules). We can get 10 O atoms on the left (to balance the 10
on the right) by putting a coefficient of 5 in front of O_{2}.
Our equation now becomes

C_{3}H_{8}(g) + 5O_{2}(g)
----------> 3CO_{2}(g) + 4H_{2}O(g)

This equation is balanced. The number of atoms of each element on both sides of the equation is as follows: 3 atoms of C, 8 atoms of H and 10 atoms of O. You should always check your final equation to be sure it really is balanced.

**Example 3**

Butane (C_{4}H_{10}) burns in oxygen (O_{2})
to form carbon dioxide (CO_{2}) and water (H_{2}O).
Write a balanced equation for this reaction showing all substances as gases.

**Solution:**

The word equation for the reaction is

butane + oxygen ----------> carbon dioxide + water

Replacing each substance with its chemical formula gives

C_{4}H_{10}(g) + O_{2}(g)
----------> CO_{2}(g) + H_{2}O(g)

Starting with carbon, we see there are 4 C atoms on the left (in C_{4}H_{10})
but only 1 C atom on the right (in CO_{2}). We can get 4
C atoms on the right by putting a coefficient of 4 in front of CO_{2}.
Our equation now becomes

C_{4}H_{10}(g) + O_{2}(g)
----------> 4CO_{2}(g) + H_{2}O(g)

Carbon is now balanced, with 4 C atoms on each side. Now let's
look at hydrogen. Thre are 10 H atoms on the left (in C_{4}H_{10})
and only 2 on the right (in H_{2}O) but we can get 10 H atoms on
the right if we put a coefficient of 5 in front of H_{2}O.
This gives

C_{4}H_{10}(g) + O_{2}(g)
----------> 4CO_{2}(g) + 5H_{2}O(g)

Now lets look at oxygen. There are only 2 O atoms on the left
(in O_{2}) but there are 13 on the right (8 of these are in CO_{2}
and 5 of these are in H_{2}O). Here, we seem to have a problem.
We have an odd number of oxygen atoms on the right side (13) but on the
left side, the oxygen atoms can only come in pairs -- that is, they can
only produce an even number . How are we going to get 13 O atoms
when they must come in pairs?

We solve this problem by doing something that might seem rather unusual.
We use a fractional coefficient. If we put a coefficient of 13 /
2 (that's thirteen halves) in front of O_{2} and write the equation
as

C_{4}H_{10}(g) + (13 / 2) O_{2}(g)
----------> 4CO_{2}(g) + 5H_{2}O(g)

This balances the equation. There are now 13 atoms of oxygen atoms
on the left side just like the right side. Recall from the notes
on Counting Atoms in Chemical Formulas that you multiply the coefficeint
by the subscript to get the number of atoms. For O_{2} this
multiplication is

__13__ x 2 = 13

2

And on the right hand side, we have

4 * 2 + 5 * 1 = 8 + 5 = 13

Although balanced, some may find it troubling that a fraction was used as a coefficient. Later, we will develope the mole concept, and from then on, there will be nothing wrong with fractional coefficients. Right now, however, we are interpreting a chemical formula to represent one molecule (or "formula unit" for ionic compounds, which don't exist as molecules). We can't have a fraction of a molecule -- they come only in whole numbers -- like people.

We can easily get an equation balanced using only integers (no fractions) by multiplying ALL the coefficients by 2. The number 2 was chosen because the denominator of our fraction is 2, and we want to "clear the fraction". Multiplying all coefficients by 2 gives

2C_{4}H_{10}(g) + 13O_{2}(g)
----------> 8CO_{2}(g) + 10H_{2}O(g)

The above equation is balanced with the smallest possible integers, that is, in lowest terms. Each element has the same number of atoms on both sides of the equation. The numbers are as follows: 8 C atoms, 20 H atoms, and 26 O atoms.

A mistake sometimes made when trying to clear the fraction is to multiply only the fractional coefficient by 2 and write

C_{4}H_{10}(g) + 13O_{2}(g)
----------> 4CO_{2}(g) + 5H_{2}O(g)

This does not balance oxygen. There are 26 O atoms on the left
(in O_{2}) and 13 O atoms on the right (8 in CO_{2} and
5 in H_{2}O). Think of the set of coefficients in a balanced
equation as the ingredients in a cooking recepie. If you want to
double the recepie, you must double ALL ingredients.

**Example 4**

Table sugar (C_{12}H_{22}O_{11}), a solid, can
be made to burn in oxygen (O_{2}), forming carbon dioxide (CO_{2})
and water vapor (H_{2}O) as products. Write a balanced equation
for this reaction, indicating the correct physical state for all substances
involved.

**Solution:**

The word equation is

sugar + oxygen ----------> carbon dioxide + water

Replacing each substance with its formula gives

C_{12}H_{22}O_{11}(s) +
O_{2}(g) ----------> CO_{2}(g)
+ H_{2}O(g)

Starting with carbon, we note that there are 12 C atoms on the left
(in C_{12}H_{22}O_{11}) but only 1 C atom on the
right (in CO_{2}). We can get 12 C atoms on the right by
putting a coefficient of 12 in front of CO_{2}. Our equation
now becomes

C_{12}H_{22}O_{11}(s) +
O_{2}(g) ----------> 12CO_{2}(g)
+ H_{2}O(g)

C atoms are now balanced, with 12 on each side. Now let's look
at hydrogen. There are 22 H atoms on the left (in C_{12}H_{22}O_{11})
but only 2 H atoms on the right (in H_{2}O). We can get 22
H atoms on the right by putting a coefficient of 11 in front of H_{2}O.
Our equation now becomes

C_{12}H_{22}O_{11}(s) +
O_{2}(g) ----------> 12CO_{2}(g)
+ 11H_{2}O(g)

Now look at oxygen. Let's consider O_{2} last, because
it's the one we're going to adjust to make everything work. We have
to take stock of what we have elsewhere, then decide what must be done
with O_{2} to bring the O atoms into balance. On the right
side, we have a total of 35 O atoms (24 of these are in CO_{2}
and 11 are in H_{2}O). We must therefore have 35 O atoms
on the left side also. We have 11 of these required 35 O atoms in
C_{12}H_{22}O_{11}. This leaves 35 - 11 or
24 O atoms that must be supplied by O_{2}. By putting a coefficient
of 12 in front of O_{2} we obtain the needed 24 O atoms.
Therefore, the balanced equation is

C_{12}H_{22}O_{11}(s) +
12O_{2}(g) ----------> 12CO_{2}(g)
+ 11H_{2}O(g)

This is the balanced equation. The number of atoms of each element -- which is the same on both sides of the equation -- is as follows: 12 C atoms, 22 H atoms, and 35 O atoms.

**Example 5**

Potassium metal, a solid, reacts with liquid bromine to form solid potassium bromide. Potassium is monatomic and bromine is diatomic. The formula of potassium bromide should be obvious from the known ion charges for these elements. Write a balanced equation to describe the reaction.

**Solution:**

The word equation is

potassium + bromine ----------> potassium bromide

Replacing the chemical names with the corresponding formulas gives

K(s) + Br_{2}(l) ---------->
KBr(s)

Note that the formula of potassium bromide was arrived at by recognizing
that potassium forms K^{+} ions and bromine forms Br^{-}
ions. In lowest terms, it takes one of each ion to give a neutral
formula. If it is not clear to you how I arrive at KBr for potassium
bromide, please review the lectures notes titled Chemical Nomenclature.

Looking at the above equation, we see that K atoms are balanced (1 K
atom on each side) but Br atoms are not. There are 2 Br atoms on
the left (in Br_{2}) but only 1 Br atom on the right (in KBr).
We can get 2 Br atoms on the right by putting a coefficient of 2 in front
of KBr. We now have

K(s) + Br_{2}(l) ---------->
2KBr(s)

Br atoms are now balanced -- with 2 on each side. But now K is no longer balanced. There is only 1 K atom on the left (in K) but 2 K atoms on the right (in KBr). We can get 2 K atoms on the left by putting a coefficient of 2 in front of K. We now have

2K(s) + Br_{2}(g) ---------->
2KBr(s)

This is the the balanced equation. There are 2 K atoms on both sides and 2 Br atoms on both sides.

**Example 6**

When aqueous solutions of potassium carbonate and calcium chloride are mixed, they produce solid calcium carbonate and aqueous potassium chloride. The formulas of the substances in the reaction should be obvious from the known ion charges. Write a balanced equation for this reaction.

**Solution:**

The word equation is

potassium carbonate + calcium chloride ----------> calcium carbonate + potassium chloride

The formulas can be figured out using the methods in the lecture notes titled Chemical Nomenclature. Replacing the chemical names with the corresponding chemical formulas gives

K_{2}CO_{3}(aq) + CaCl_{2}(aq)
----------> CaCO_{3}(s) + KCl(aq)

Notice that the carbonate group (CO_{3}^{2-}, except
that the charge is not written in the formula because it is cancelled by
the cation charge) remains intact during the reaction. Therefore,
rather than balancing C atoms and O atoms separately, we can save some
time (and steps) by balancing carbonate as a group. As of right now,
carbonate is balanced. There is 1 carbonate group on each side of
the equation. If carbonate becomes unbalanced during the course of
our efforts balancing the other atoms, we will have to deal with it later.
For now, let's turn our attention to another atom. Let's look at
potassium. We have 2 K atoms on the left side (in K_{2}CO_{3})
but only one K atom on the right side (in KCl). We can get 2 K atoms
on the right side by putting a coefficient of 2 in front of KCl.
We now have

K_{2}CO_{3}(aq) + CaCl_{2}(aq)
----------> CaCO_{3}(s) + 2KCl(aq)

All atoms are now balanced. Sometimes we have those lucky breaks
where balancing one atom balances others at the same time. If you
will look back at the original equation, you will see that Cl atoms were
not balanced (2 on the left and 1 on the right). The balancing step
we just carried out was performed with K atoms in mind, but it has balanced
Cl atoms as well. Ca atoms and the CO_{3}^{2-} were
balanced to begin with, and nothing we did while balancing K (and Cl) affected
that balance, so Ca and CO_{3}^{2-} remain balanced.
On at atom by atom basis, we have the following number of atoms of each
element on both sides of the equation: 2 K atoms, 1 C atom, 3 O atoms,
and 2 Cl atoms.

**Example 7**

Ferric oxide, a solid, reacts with carbon monoxide, a gas, to form solid iron metal and carbon dioxide gas. Iron is a monatomic element. The formulas of the other substances in this reaction can be determined from their chemical names. Write a balanced equation for this reaction.

**Solution:**

The word equation is

ferric oxide + carbon monoxide ----------> iron + carbon dioxide

Since iron forms the Fe^{2+} and Fe^{3+} ions, the ferric
ion must be Fe^{3+}, because the suffix -ic refers to the ion having
the higher charge. Since the oxide ion is O^{2-}, the formula
must be Fe_{2}O_{3}. If it not clear how I obtained
this formula, please study the Chemical Nomenclature notes. The Greek
prefix mono- means 1 and di- means 2. Therefore carbon monoxide is
CO and carbon dioxide is CO_{2}. (If no prefix is given for
the first element, it is mono- by default). Having figured out all
the formulas, we can now write our first draft of the equation.

Fe_{2}O_{3}(s) + CO(g)
----------> Fe(s) + CO_{2}(g)

This equation is rather difficult to balance on an atom by atom basis. It tends to lead one down a twisted path. What would seem like a logical start would be to note that C is already balanced and that putting a 2 in front of Fe will balance Fe.

Fe_{2}O_{3}(s) + CO(g)
----------> 2Fe(s) + CO_{2}(g)

The problem occurs when you try to balance oxygen. There are 4
O atoms on the left (3 in Fe_{2}O_{3} and one in CO).
There are 2 O atoms on the right, in CO_{2}. If you
put a 2 in front of CO_{2} to get 4 O atoms on the right like this:

Fe_{2}O_{3}(s) + CO(g)
----------> 2Fe(s) + 2CO_{2}(g)

you will have messed up the balance in C atoms. If you then put a 2 in front of CO to restore the C atom balance like this:

Fe_{2}O_{3}(s) + 2CO(g)
----------> 2Fe(s) + 2CO_{2}(g)

you've messed up the O atom balance again. Situations like this can drive you crazy!

The trick to seeing your way clearly in getting this equation balanced
is to "get the big picture". Let's think about what really happens
in this reaction. The CO becomes CO_{2} which means each
CO molecule that participates in this reaction must pick up 1 O atom.
The Fe_{2}O_{3} is stripped of all its O atoms, to become
just Fe. So each formula unit of Fe_{2}O_{3} that
participates in this reaction must lose 3 O atoms. Every O atom that
is lost by Fe_{2}O_{3} must be picked up by a CO molecule,
because there is no monatomic oxygen (O) being formed in this reaction.

How can each Fe_{2}O_{3} formula unit lose 3 O atoms,
and each CO molecule gain only one O atom, and yet not have any O atoms
left over? There must be 3 molecules of CO for every 1 formula unit
of Fe_{2}O_{3} in the reaction. That way, all the
O atoms lost by the Fe_{2}O_{3} have somewhere to go, and
are not "left out in the cold". Of course, those 3 CO molecules will
become CO_{2} molecules when they receive the O atoms, so 3 CO_{2}
molecules will be formed. The balanced equation is therefore

Fe_{2}O_{3}(s) + 3CO(g)
----------> 2Fe(s) + 3CO_{2}(g)

All atoms are now balanced. On both sides of the equation, there are 2 Fe atoms, 3 C atoms, and 6 O atoms.

I hope you have found these notes helpful. Your questions, comments and suggestions for improvement are always welcome.

SUGGESTED READING

Section 3.7, "Writing and Balancing Chemical Equations", pages 100 - 104, in your Hill & Petrucci General Chemistry textbook.

*This page was last modified Wednesday October 13, 1999*

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