Balancing Chemical Equations - Concept

Alright so let's talk about the balancing of chemical equations when we're dealing with chemical equations and balancing them, we have to make sure we've adhered to a simple rule and that's the conservation of matter. You might know it as a conservation of mass and that is saying that matter is not created or destroyed it's only changed. Okay so we want to make sure that we don't like magically make up new elements as we're dealing with chemical equations. We're going to make sure whatever we have on the reactive side is exactly the number of particles we have on the product side. And so let's talk about what, let's put into action what I'm talking about, let's deal with this, we have this written chemical equation and I'm going to put that into a skeletal equation or the actual chemicals. So we have liquid carbon disulphide.

Okay so that's CS2 it's a liquid so we have to denote that, reacts with oxygen gas and oxygen gas is one of our diatomics so we're going to say it's O2 and it's a gas, yields or produces carbon dioxide gas and sulfur dioxide gas. Okay great but we want to make sure when we're dealing with chemical equations that the number of particles are the number of elements that you have on the reacting side equals the number of elements you have on the product side. Let's go ahead and check that we have carbon, sulfur and oxygen here which means we should have strictly a carbon, sulfur and oxygen on the product side. Okay and we want to make sure when we're dealing with these that we do not change the actual molecules themselves, we only change the number of molecules that we have.

The only place I actually want to put numbers and make changes is the place before that, we call those coefficients. Okay so on the reacting side we have 1 carbon, we have 2 sulfur atoms and we have 2 oxygen atoms. On the product side we have 1 carbon, 1 sulfur and 2+2 is 4 oxygens okay great. So now tight now before we even balance we notice that we have an even number of elements on either side to make sure they're even and balanced. So right now the carbons are balanced, we have 1 carbon on the reactive side, 1 on the product side so we're good there.

Looking at sulfur we have 2 sulfur atoms and we have 1 sulfur atom over here so we have to change that. The place to change it is before the compound right here, so I have 2 on that side I want to make 2 on this side. So I'm going to put a 2 here okay, that tells me I produced 2 sulfur dioxide particles. So 2 sulfurs but then I also changed the oxygen. So we have 4 now plus that other 2 which is 6 okay. So now our carbons are balanced check our sulfurs are balanced check. We want to make sure our oxygen that are balanced. We have 2 on the reacting side and 6 on the product side. I'm going to change that, so over here we want to say there's 3, we need 3 oxygen gas particles to react with one carbon disulfide particle. So we're going to change that to 6 and now 1 2 6, 1 2 6 we're good to go this is completely balanced great reaction.

Okay let's do something a little more complicated going over here before we dive into this equation. Its reaction we have on the board which is potassium chromate plus lead 2 nitrate yield potassium nitrate plus lead 2 pomade. We want to make sure we balance this properly but there's a lot of atoms in this. So we're going to make sure this is easy on ourselves before we jump right into it we should notice a pattern. We have chromate molecule, polyatomic ion here, we also have a chromate polyatomic ion here I can just keep those together. I also have a nitrate particle polyatomic ion here and a nitrate here. I don't have to separate the nitrogen and the oxygen I can keep them together since they're on the reacting and product side both. So I'm going to separate everything else out potassium, chromate and nitrate.

And I'm going to do the same thing on the product side, okay so I have 2 potassium, 1 chromate, 1 lead and 2 nitrates. I have 1 potassium, 1 chromate, 1 lead and 1 nitrate. Okay so let's deal with the first one first, potassium we have 2 over here, we have 1 over here I do not want to put 2 to part with the compound itself I do not want to change it at all. I want to put it in front so I'm going to make that 2. So everyone has now 2 potassiums over here but that also changed my nitrate to 2 and let's see so we have 2 1 1 2 1 1 2 it looks like the whole thing is done we're good to go. Let's look at something a little bit more complicated. Let's look at this reaction here, so we have C3H6 reacts with oxygen gas to produce CO2+H2O.

Now let's do what we've been doing, writing our elements out C, H and L, C, H and L okay we have 3 carbons, we have 6 hydrogens and 2 oxygens, we have 1 carbon, we have 2 oxygens and we have 3 I'm sorry 2 hydrogens, 3 oxygens. Okay so first things first, we're going to balance our carbons, so we have 3 on the reacting side, 1 on the product side I want to put a 3 in front let's change this to 3, that changes this to 6 plus 1 is 7. Okay hydrogens have 6 on the reacting side 2 on the product side, so I want to put a 3 here making this 6 that also changes our oxygen so we now have 3+6 which is 9. Okay so we get to our oxygen. Oxygens we have 2 on the reacting side and 9 on the product side.

How are we going to make this change to 9? Well alright so we'll put a 4.5 there, 4.5 times 2 is 9 right but that looks terrible. We don't ever want to put a decimal or a fraction as a coefficient. So how are we going to get rid of that? Well a good solution is to multiply the whole reaction by 2, so okay that means I'm going to change all my coefficients to multiply them by 2. So I then I'm going to where should I put this? I'll just put it here, I then I'm going to make this 2 because 1 times 2 is 2, 4.5 times 2 is 9, 3 times 2 is 6, so we're going to change that to a 6 and 3 times 2 is 6. So then let's check it, we have 6 carbons, 6 carbons check, we have 12 hydrogens, 12 hydrogens awesome, we now have 18 oxygens, we now have 12+6 great 18 oxygens we're done and that is how you balance chemical reactions.