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Cara Menyetarakan Reaksi ( 3 )

Example 2:
__B2 + __W2 --> __ BW3. Underneath B2 is one molecule made from two overlapping blue atoms. Underneath W2 is one molecule made from two overlapping white atoms. Underneath BW3 is one molecule made from 1 blue atom with three white atoms attached to it.
Again, we drew the molecules represented by the formulas in the equation. First let's balance the blue atoms. How many blue atoms are there on the left, or "reactants side" of the equation? How many on the right, or "products side"? In order to get one more atom of blue on the right side, what do we need to add? Yes, we need to add an entire group. So let's do that.
__B2 + __W2 ' 2 BW3. Underneath B2 is one diatomic blue molecule. Underneath W2 is one diatomic white molecule. Underneath 2 BW3 are two molecules, each with one central blue atom with three white atoms attached.
See that we have also added the red coefficient, 2, in front of the formula, BW3 , to reflect the addition. This now balances our blue atoms, but the whites are still unbalanced. How many white atoms do we have on the right side of the equation? [Answer: 6] And how do we get 6 atoms of white on the left hand side? Yes, by adding groups. What is the total number of groups we need on the left to balance the 6 whites on the right?
__B2 + 3 W2 ' 2 BW3. Underneath B2 is one diatomic blue molecule. Underneath 3 W2 are three diatomic white molecules. Underneath 2 BW3 are two molecules, each with one central blue atom with three white atoms attached.
So now we have another correctly balanced equation.
B2 + 3 W2 --> 2 BW3
Example 3:
__Y + __R2 --> __ Y2R3. Underneath Y is one yellow atom. Underneath R2 is one diatomic red molecule. Underneath Y2R3 is one group of atoms, which can be either a molecule or a unit salt formula, made from 2 yellow atoms connected with three red atoms.

With this equation, first let's look at the yellow atoms. There is 1 yellow on the left, but 2 yellows on the right. What do we do?
2 Y + __ R2 --> __ Y2R3. Underneath 2 Y are two single yellow atoms. Underneath R2 is one diatomic red molecule. Underneath Y2R3 is one group of atoms with 2 yellow and 3 red atoms connected together.
OK. This balances our yellows, but our reds are still unbalanced. What do we do next?
Construction Caution sign
A word of caution here. Many students at this point make a BIG mistake! They actually take atoms AWAY from Y2R3 and change the formula to Y2R2, as shown below! Don't you do this!
Two boxes demonstrating that it is against the rules of balancing equations to take atoms away from a formula by changing the subscripts in order to balance atoms in an equation. The examples are Y2R3 is NOT equal to Y2R2. Underneath Y2R3 is a group of atoms made from 2 yellow and 3 red atoms connected together. Under Y2R2 is a group of two yellow and two red atoms connected together, but there are question mark signs pointing to the subscript of '2' behind R in Y2R2. And there is another question mark pointing to the place where the third red atom is missing from the drawing.
The ONLY thing we can EVER do in balancing equations, once the formulas are correct, is to ADD groups.
So what groups should we add? If we add one group of R2, as shown below, this still doesn't balance the reds.
2 Y + 2 R2 ' __ Y2R3. Under 2 Y are two single yellow atoms. Under 2 R2 are two red diatomic molecules (in other words, two pairs of overlapping atoms to give 4 atoms total but in two different molecules). Under Y2R3 is one group of 2 yellow and 3 red atoms connected together. There is also a large purple question mark, begging the question about what we do next to balance this equation.
Can you guess the secret? The secret is to find the least common multiple, (yes, an application of math!) between the original 2 reds on the left side and the 3 reds on the right. What is the least common multiple of 2 and 3? [Answer: 6] How do we get 6 reds on both sides?
2 Y + 3 R2 ' 2 Y2R3. Under 2 Y are two single yellow atoms. Under 3 R2 are 3 red diatomic molecules. Under 2 Y2R3 are two groups of atoms, each group containing 2 yellow atoms and 3 red atoms connected together.
Yes, "3 groups of 2" and "2 groups of 3."
But now the yellows are unbalanced again. What do we do next? Remember, 4 Y does NOT equal Y4.
Reminder boxes. Two boxes with a 'do not equal' sign in between. The first box has 4 Y written in it. Below the 4 Y are 4 separate single yellow atoms. The second box has the formula Y4. Below Y4 is one molecule made from 4 yellow atoms connected together. These two formulas are not equivalent.
We change the coefficient in front of "Y" to 4.
4 Y + 3 R2 ' 2 Y2R3. Under 4 Y are 4 separate single yellow atoms. Under 3 R2 are 3 red diatomic molecules. Under 2 Y2R3 are two groups of atoms, each one with 2 yellow and 3 red atoms connected together.
So the final balanced equation is:
4 Y + 3 R2 --> 2 Y2R3
One last word: When balancing equations, you ALWAYS want the lowest possible numbers. For example, the above equation may also be written as:
8 Y + 6 R2 ' 4 Y2R3. 12 Y + 9 R2 ' 6 Y2R3. 24 Y + 18 R2 ' 12 Y2R3. It is obvious that these equations are multiples of each other. The balanced equation must always have the lowest coefficients possible and still have a balanced equation.

All of these equations are also technically "balanced," but on a test ONLY the lowest numbered choice, i.e., "4Y + 3R2 => 2Y2R3" would be correct. We ALWAYS want the equations with the lowest numbers.
Review:
You should now understand the basics for balancing chemical equations.
1. ALL balanced equations must obey the Law of Conservation of Mass, which means that they must have thesame number of atoms of each kind on both sides of the equation.
2. The subscript tells how many atoms of each kind there are in a formula.
3. The coefficient is a multiplier, multiplies every atom in the formula along with its subscript, and is the ONLYnumber which may be changed in balancing equations.(Once formulas are correct, the ONLY way to add atoms is by adding groups.)
4. P.S. You ALWAYS want the lowest possible numbers.


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