page 3 - more complex equations
We will now take a look at more complex equations; still using the same concepts of the last few pages.
Take a look at the following equation:
Pb(No3)2 => PbO + NO2 + O2
When you are first confronted by this type of equations, it's always a great idea to work from left to right.
A technique here is to leave the atom(s) that appears most on the right to last; in this case it's the oxygen.
Let's make a start with the Nitrogen on the left hand side there is 2, on the right there is 1, so we will correct it.
Pb(No3)2 => PbO + 2NO2 + O2
Now let's us count the oxygens, there are 6 on the left, but 7 on the right. In this scenario you have two choices. The first applies when you have oxygen (O2) or hydrogen (H2) on one side of the equation; here we have an O2.
The easiest way to look at this is just lose an oxygen from the right, which will mean we place a half (1/2) in front of the O2.
Pb(No3)2 => PbO + NO2 + 1/2O2
This techniques is frowned upon in chemistry; however, this is only one of the solutions. But if you don't like this solution 2 is doubling everything, like so:
2Pb(No3)2 => 2PbO + 4NO2 + O2
Yes, it can be slightly unusual having halves in some equations, but sometimes you don't have a choice, so just add them, it's ok.
Take a look at the following equation:
Pb(No3)2 => PbO + NO2 + O2
When you are first confronted by this type of equations, it's always a great idea to work from left to right.
A technique here is to leave the atom(s) that appears most on the right to last; in this case it's the oxygen.
Let's make a start with the Nitrogen on the left hand side there is 2, on the right there is 1, so we will correct it.
Pb(No3)2 => PbO + 2NO2 + O2
Now let's us count the oxygens, there are 6 on the left, but 7 on the right. In this scenario you have two choices. The first applies when you have oxygen (O2) or hydrogen (H2) on one side of the equation; here we have an O2.
The easiest way to look at this is just lose an oxygen from the right, which will mean we place a half (1/2) in front of the O2.
Pb(No3)2 => PbO + NO2 + 1/2O2
This techniques is frowned upon in chemistry; however, this is only one of the solutions. But if you don't like this solution 2 is doubling everything, like so:
2Pb(No3)2 => 2PbO + 4NO2 + O2
Yes, it can be slightly unusual having halves in some equations, but sometimes you don't have a choice, so just add them, it's ok.
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