High School Chemistry : Balancing Chemical Equations
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Example Question #271 : High School Chemistry
Balance the following equation:
When balancing a chemical equation, it means that we are adjusting the coefficients in front of every compound so that the number of atoms for every element will match how many will be on the other side of the arrow.
Initially, we must take "inventory" of how many atoms of each element we have on both sides of the arrow.
On the left side:
1 iron, 2 chlorine
On the right side:
1 iron, 3 chlorine
While the iron atoms are balanced, the chlorine atoms are not. Now we must think, what is the least common multiple of 2 and 3? We can quickly realize it is 6. So we must add a coefficient in front of every term that includes a chlorine so we may make it 6 chlorine atoms on both sides of the arrow. (The coefficient will be multiplied by the subscript attached to the chlorine atom.)
While the chlorine atoms are balanced now, we have lost the balance for the iron atoms. This can be quickly resolved. On the left, we still only have one iron atom. On the right, it has changed to 2 iron atoms. This means that in order to balance the number of iron atoms, we just need to add a 2 coefficient in front of the iron on the left side of the arrow.
Now let's do one last tally to make sure we've balanced everything.
On the left:
2 iron, 6 chlorine
On the right:
2 iron, 6 chlorine
The equation has been balanced.
Example Question #272 : High School Chemistry
Balance the following equation:
When balancing a chemical equation, it means that we are adjusting the coefficients in front of every compound so that the number of atoms for every element will match how many will be on the other side of the arrow.
Initially, we must take "inventory" of how many atoms of each element we have on both sides of the arrow.
On the left side:
1 iron, 3 bromine, 2 hydrogen, 1 sulfur, 4 O
On the right side:
2 iron, 1 bromine, 1 hydrogen, 3 sulfur, 12 O
Since there are quite a few components in this example, all of which need balancing, we just need to choose one element to start with and continue from there. Let's look at iron. Since there is only one iron atom on the left side, and 2 on the right side, let's think about what the lowest common multiple between 1 and 2 is - we can quickly decide that it is 2. This means we may add a 2 coefficient in front of the 
Adding the 2 coefficient in front of 
By placing that 6, we change the number of hydrogen atoms on the right side to 6 from 1. Comparing that to number of hydrogen atoms on the left side of the arrow (2), we can determine that the lowest common multiple between 6 and 2 is 6. This means that we must place a 3 coefficient in front of the 
By placing that last coefficient, this affected the number of sulfur and oxygen atoms on the left side - we now have 3 sulfur atoms and 12 oxygen atoms on the left side. Comparing this to the right side, where we also have 3 sulfur atoms and 12 oxygen atoms. So this last step has balanced the rest of our atoms!
Now let's do one last tally to make sure we've balanced everything.
On the left:
2 iron, 6 bromine, 6 hydrogen, 3 sulfur and 12 oxygen
On the right:
2 iron, 6 bromine, 6 hydrogen, 3 sulfur and 12 oxygen
The equation has been balanced.
Example Question #21 : Balancing Chemical Equations
Balance the following equation:
When balancing a chemical equation, it means that we are adjusting the coefficients in front of every compound so that the number of atoms for every element will match how many will be on the other side of the arrow.
Initially, we must take "inventory" of how many atoms of each element we have on both sides of the arrow.
On the left side:
1 potassium, 5 oxygen, 4 hydrogen, 1 P
On the right side:
3 potassium, 5 oxygen, 2 hydrogen, 1 P
Since there are quite a few components in this example, all of which need balancing, we just need to choose one element to start with and continue from there. Let's look at potassium. Since there is only one potassium atom on the left side, and 3 on the right side, let's think about what the lowest common multiple between 1 and 3 is - we can quickly decide that it is 3. This means we may place a 3 coefficient in front of the 
Adding the 3 coefficient in front of 
This brings our total of oxygen atoms on the right side to 7 as well. So we have indirectly balanced the oxygen atoms. Since the only element left to balance is phosphorus, we realize that phosphorus on both sides is already balanced. We have balanced our equation.
Now let's do one last tally to make sure we've balanced everything.
On the left:
3 potassium, 7 oxygen, 6 hydrogen, 1 phosphorus
On the right:
3 potassium, 7 oxygen, 6 hydrogen, 1 phosphorus
The equation has been balanced.
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