[Music] all right so let's talk about the balancing of chemical equations and we're dealing with chemical equations and balancing them we have to make sure we adhere to a simple rule and that's a 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 going as we're dealing with chemical equations we want to make sure whatever we have on the reactant side is exactly the number of particles we have on the product side and so let's talk about what I'm let's put into action what I'm talking about let's deal with this we have this written chemical equation um and I'm going to put that into a skeletal equation or um the actual chemicals so we have liquid carbon disulfide okay so that's CS2 we're it's a liquid so we have to denote that uh 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 or the number of elements that you have on the reactant side equals the number of elements you have on the product side so let's go ahead and check that we have carbon sulfur and oxygen here which means we should have strictly 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 so the only place I actually want to put number and make changes is the place before that we call those coefficients okay so on the product or sorry on the reactant side we have one carbon we have two sulfur atoms and we have two oxygen atoms on the product side we have one carbon uh one sulfur and 2 plus 2 is four oxygens okay great so now right now before we even balanced it we notice that we have an uneven uh number of elements on either side so make sure they're even in balanced so right now the carbons are balanced we have one carbon on the reactant side one on the product side so we're good there looking at sulfur we have two sufur atoms and we have one sulfur atom over here so we have to change that the place we change it is before the before the element or sorry before the compound right here so I have two on that side I want to make two on this side so I'm going to put a two here okay that tells me I produced two sulfur dioxide particles so two sulfur but then I also changed the oxygen so we have four now plus the other two which is six okay so now our carbons are balanced check our sulfur are balanced check we want to make sure our oxygen then are balanced we have two on the reactant side and six on the product side we going to change that so over here we want to say there's three we need three oxygen gas particles to react with one carbon carbon disulfide particle so we're going to change that to six and now it's one 2 six one 2 six we are good to go this is completely balanced great reaction okay let's do something a little bit more complicated going over here before we dive into this equation this reaction we have on the board which is potassium chromate uh plus lead to nitrate yields pottassium nitrate plus lead to chromate um we want to make sure we balance this properly but it's there's a lot of molecules or sorry a lot of atoms in this so we want to make sure this easy on ourselves before we jump right into it we should notice a pattern um we have a chromate molecule or 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 reactants and product side both so I'm going to separate everything else out potassium chromate lead and nitrate oops and I'm going do the same thing on the product side uh okay um okay so I have two potassiums uh one chromate one lead and two nitrates I have one potassium one chromate one lead and one nitrate okay so let's deal with the first one first potassium we have two over here we have one over here I do not want to put a two within the part within the um compound itself I do not want to change this at all I want to put it in front so I'm going to make that two so I'm have now have two potassiums over here but that also changes my nrat to two and um let's see so we have 2 one one2 21 one2 looks like the whole thing's done we are good to go let's look at something a little bit more complicated let's look at um this part this uh reaction here so we have C3 H6 reacts with oxygen gas to produce CO2 plus H2O all right let's see what we've been doing writing our elements out c h and O CH H and O okay we have three carbons we have six hydrogens and two oxygens we have one carbon we have two oxygens and we have three oh sorry two hydrogen's three oxygen okay so first things first we want to balance our carbons so you have three on the reactant side one on the product side I want to put a three in front so that changes this to three that changes this to 6 + 1 is 7 okay hydrogens I have six on the reactant side two on the product side so I want to put a three here making this six but it also changes our oxygen so we now have 3 + uh 6 which is n okay so me go to our oxygen our oxygen we have two on the reactant side and nine on the product side how are we going to make this um change to nine well all right so we'll put a 4.5 there 4.5 * 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 um well good solution is to multiply the whole reaction by two so okay that means I'm going to change all my coefficients to multiply by two so I then I'm going to where should I put this I'll just put it here and then I'm going to make this two because 1 * 2 is 2 4.5 * 2 is 9 3 * 2 is six so we're going to change that to a six and 3 * uh 2 is 6 so then let's check it we have six carbons six carbons check we have 12 hydrogens 12 hydrogens awesome we now have 18 oxygens we now have 12 + 6 great 18 oxygen we are done and that is how you balance chemical reactions