Balancing Equations
Why do we balance equations? Before we start, get this down first. Now let's take a simple chemical equation, unbalanced:
Cu + O2 -> CuO Now this only tells us when we react Copper and Oxygen gas together, we will get Copper (II) Oxide. But what if scientists want to know how much of something we will get at the end instead of just what they will get at the end? This is where scientists take advantage of the Law of Conservation of Mass and predict how much of something we need or will get out of a chemical equation. Basically, the objective is to make sure both sides has the same amount of stuff.
Easy?
In the example above, we have 1 unit (as you see in the "ratios" chapter, this unit can be anything, from atoms to molecules to moles to grams) of Copper and 2 units of oxygen gas. This is not the same with the right, which has 1 less unit of oxygen. So we balance it. How?
Here's a hint: the answer is
2 Cu + O2 -> 2 CuO
How did you do that?
Balancing equations is something that really isn't very hard once you understand how to do it.
Before we start, let's start with the fundamentals: why do we have to balance equations? Because chemical equations must have the same amount of each type of atom on each side.
Chemical equations come in this form:
A+B -> C
Remember that there is no =, only ->, and the left hand side, the amount of reactants, must equal to the amount of products and the right hand side.
To explain how to balance equations, let's start with a example. (This equation is not WRONG, it just isn't balanced. These are usually called skeleton equations.)
H2+O2 -> H2O
Notice the 2 in H2O is a subscript (written below the symbol), but when balancing equation you can ONLY add coefficients, which is written on the same level but before the symbol (to its left).
Now, the equation is obviously NOT balanced because there are 2 oxygen atoms on the left but only one on the right. So how do you balance it?
The key thing to remember here is to add coefficients until both sides are equal in amounts.
The easiest way to start with is to write down the list of reactants and products for both sides.
H2+O2 -> H2O
H 2 H 2
O 2 O
As you can see, hydrogen is already balanced because there is two of each (shown by H 2) on both sides. Oxygen, however, is not.
To balance oxygen, you should multiply it by two, to balance it with the other side.
HOWEVER that means the hydrogen on the right is also affect too, as H2O is a compund, and you cannot place coefficients in the middle of the formula. (i.e. 2H2O is correct, while H22O is a big mistake).
H2+O2 -> 2H2O
H 2 2 H 2 = 4 H
O 2 2 O
That would mean there is not 4 hydrogen atoms on the right side - the equation is still unbalanced!
Now by observation you should see that by multiplying the H on the left hand side by 2 the number will increase to 4, causing the amount of hydrogen to be equal and therefore balancing the equation. Since hydrogen is on its own, changing its coefficient will only affect itself.
2 H2+O2 -> 2 H2O
2 H 2 2 H 2
O 2 2 O
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Let's try another example.
H2 + Cl2 ---> HCl
Again, list everything out first. Once you get good at this, it will become optional.
H 2 H
Cl 2 Cl
This is an easy one. You need two more of H and Cl on the right side to satisfy and balance out with the left side, so you simply have to add 2 to the right hand side. Clear?
H2 + Cl2 ---> 2 HCl
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O2 -> O3
This is a bit trickier but it highlights an important concept in balancing equations: finding the LCM. Remember all coefficients must be whole numbers, so a simple 2/3 will not do.
There are two ways to approach such a question.
The first one is to see that a coefficient of 3/2 will balance this equation, so to convert 2/3 to a whole number, you need to multiple both sides by 2 to eliminate the denominator.
See it in action here:
3/2 O2 -> O3
This theoretically works as after cancelling out we'd be left with O3 -> O3 which would give a balanced equation. However, since fractional coefficients cannot exist, one must convert the fractional to an interger.
3 O2 -> 2O3
Simply multiply both sides by two.
If you are good at maths, then you should see that the LCM of 2 and 3 is six, so you should be able to instantly fill in the correct coefficients of both.
But both approaches work well.
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S8 + F2 ---> SF6
A bit harder!
A good rule of thumb is to start with higher quantities first so you won't have to make large adjustments involving LCMs later on.
In this casse add an 8 to SF6 first. Now S should be balanced, while F is not. However, simply multiplying F2 by 24 should balance both out by giving both sides 48 F atoms.
S8 + 24F2 --> 8SF6
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HARD ONE COMING THROUGH!
C2H6 + O2 ---> CO2 + H2O
Here, start with sticking a 2 in front of CO2, right hand side. This balances the C (carbon).
A 3 in front of H2O, on the right, also balances H (hydrogen) out.
C2H6 + O2 ---> 2CO2 + 3H2O
The only thing left is oxygen. Now, if you count closely, you can see there is only 2 O on the left, but 7 O on the right. Oh dear!
Don't panic! Remember, you can always use fractional coefficients during balancing as long as they become intergers at the end. So fire away, unleash those fractions!
C2H6 + (7/2)O2 ---> 2CO2 + 3H2O
Since 7/2*2=7, this fractional coefficient works.
Now the only step left is to remove those denominators, which is to multiply the whole thing by 2.
2C2H6 + 7O2 ---> 4CO2 + 6H2O
Tada!
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Important notes!
1) Sometimes you get equations that are already balanced! They are meant to trick you, so don't get fooled! All you have to do is just to leave it alone (or do what the question tells you to).
2) All coefficients must be intergers!
3) All coefficients must go before the formula, and always to its left! Never to its middle or right!
4) Definition clarification:
Equations: A+B -> C
Formula: H2O, O2, CaSiO3
That's it! Good luck! (Skill comes with experience!)
