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Coefficients are Multipliers

Now that you understand how subscripts are used in chemical formulas, we will look at how coefficients are used in chemical equations. Quite simply, a coefficient is a multiplier.
Here are some examples.
Three boxes containing respectively 1, 2 and 3 identical molecules made of 1 blue atom and 2 white atoms. The first box contains one molecule. Its formula is BW2. The second box contains 2 molecules. Its formula is 2 BW2. The third box contains 3 molecules. Its formula is 3 BW2.
The coefficient multiplies, and applies to, the ENTIRE FORMULA written after it, not just the first letter. When you have the balanced equation, you can simply multiply the coefficient by the subscript for each atom represented in the formula to find out how many total atoms you have of each kind. (Remember, when there is only one atom of a kind in the formula there is no subscript written, so we use "1" as the subscript multiplier.)
Analysis of the previous graphic, emphasizing the use of coefficients. Each of the three representations, BW2, 2 BW2 and 3 BW2 are analyzed for their total atoms and how the coefficient multiplies the number of atoms in each formula to give the total number of atoms in the box. First box: BW2 Coefficient of 1 x 1 blue atom in the molecule = 1 total blue atoms in box. Coefficient of 1 x 2 white atoms in the molecule = 2 total white atoms in box. Second Box: 2 BW2 Coefficient of 2 x 1 blue atom in each molecule = 2 total blue atoms in box. Coefficient of 2 x 2 white atoms in each molecule = 4 total atoms of white in box. Third Box: 3 BW2 Coefficient of 3 x 1 blue atom in each molecule = 3 blue atoms total in box. Coefficient of 3 x 2 white atoms in each molecule = 6 white atoms total in box.
Do not make this mistake.
Do NOT think that 3 BW3 is the same as B3W6. Even though the total number of atoms of each color is the same in each box, in the first we have 3 groups of atoms or three molecules, and in the second we have one large molecule. In real life, these two kinds of molecules would act very differently. So do NOT make the mistake of writing 3 BW3 as B3W6 or vice versa.
This is just one example of a mistake many students make. The number of atoms of each kind is the same in both cases, but these two molecules are VERY different from each other. Do not EVER make the mistake of writing "3 BW2" as "B3W6" or anything like that.

Balancing Equations.

OK. Now that you have the basics down, let's start actually balancing equations. Just remember that once all the formulas in the initial equation are correct, the ONLY thing you can do to balance an equation is to add groups by changing the coefficients. Once the formulas are correct, you must NOT change the subscripts.

Example 1:
Look at this simple equation. Immediately underneath it are drawings of the molecules these formulas represent. Our task is to find the lowest number of groups of each formula such that all the atoms are accounted for and balanced on both sides of the equation. The "reactants side" of the equation is anything written BEFORE the arrow. The"products side" of the equation is anything written AFTER the arrow.
Color code chemical equation: __R2 + __W2 --> __RW2. Blank lines are placed in front of the formulas in the equation to allow for the writing of coefficients, once the equation is balanced. The first two terms before the arrow are on the 'Reactants' side. The term after the arrow ison the 'Products' side.
....Diatomic red molecle + diatomic white molecule --> molecule made from 1 red and 2 white atoms. This is a drawing of the unbalanced equation written above.
It can be seen by inspection that there are 2 red atoms on the left and only 1 red atom on the right. Thus, this equation is NOT balanced.
In order to balance the quation, we need at least 1 more red atom on the right side, but we cannot add JUST 1 redatom. Rather, we must add an entire GROUP of atoms which contains our red atom of interest. It's somewhat like buying a box of crayons. In order to get one crayon of a certain color, you must buy the entire box, because they just don't come one crayon at a time.
We have to add at least one red atom to the right side, but in order to do that, we have to add one entire group, so let's do that and see what we get.
__R2 + ___ W2 ' 2 RW2. Below this equation are drawings of the molecules represented so far. We have 1 red diatomic molecule, 1 diatomic white molecule and 2 molecules of RW2.
Adding the group balances our red atoms, giving us 2 red atoms on each side. However, now the whites are unbalanced. We have 2 white atoms on the left side of the arrow but 4 on the right side. What should we do? Of course, "add a group."
__R2 + 2 W2 ' 2 RW2. Below are drawings of the molecules represented so far. We have 1 red diatomic molecule, 2 white diatomic molecules and 2 molecules of RW2.
Now if we look at the equation, we see that there are the same number of each kind of atom on both sides of the equation. So this equation is now balanced. All that is left for us to do is write down the coefficients.
Remember, you cannot represent "2 W2" as "W4."
This is a reminder. 2 W2 is NOT equal to W4. Below this statement are two boxes. The first has 2 white diatomic molecules. The second has 1 molecule with 4 white atoms. There is a 'not equal' sign in between the two boxes.

Nor can you represent "2 RW2" as "R2W4."
2 RW2 is NOT equal to R2W4. Below this statement are two boxes. The first has 2 molecules, each molecule with 1 red and 2 white atoms. The second box has only one molecule with 2 red and 4 white atoms, all joined in a letter 'H' shape.
So the balanced equation is:
The balanced equation: R2 + 2 W2 --> 2 RW2

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