If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:4:58

all right now we have another combustion reaction instead of instead of ethylene we now have ethane c2h6 has two carbons and six hydrogen atoms in each molecule of ethane and it is reacting its ethane gas it's reacting with molecular oxygen in gaseous form and they combust to form carbon dioxide gas and liquid water and like we've seen in previous examples this chemical equation is not balanced how can we tell well here on the left hand side we have two carbons here on the right hand side we have one carbon here on the left hand side we have six hydrogen's here on the right hand side we only have two hydrogen's here on the left hand side we have two oxygens on the right hand side we have two plus three oxygens so all of the out none of the elements here are balanced but like we did in the example with ethylene whenever you see this where you have you know several you know somewhat complex molecules involved it's good to save the element that is in a molecule by itself for last because you can just tweak this to change the number of oxygens without having any other side effects on the number of carbons or hydrogen's so what I'm going to do is I'm going to first balance like we've done before the carbons and the hydrogen's which is which are going to have implications on the oxygens because if I change the number here it's gonna change the number of oxygens if I change the number here it's gonna change the number of oxygens but lucky for me I have this dioxygen molecule on the left-hand side that I can just tweak at the end to balance the entire chemical equation so let's start with you know last time we started with carbon let's start with hydrogen this time just for kicks so over here I have six hydrogen's on the left hand side or the entire left hand side only have six hydrogen atoms on the right hand side I only have two right now so if I want to have if I want to have six I would multiply these two by three so now I have three water molecules each of them have two hydrogen atoms so I'm going to have six six hydrogen atoms on the right hand side fair enough now let's go to the carbon remember I'm saving oxygen for last carbon on the left-hand side I have two carbons how many carbons do I have on the right hand side well right now I only have one but I can change that very easily instead of having one molecule of carbon dioxide I can have two molecules of carbon dioxide and so now my carbons are balanced two carbons two carbons and now let's go to the oxygens so right now right now on I'll do this in a move color alright now on the left hand side I have two oxygens but on the right hand side what do I have let's see F two times two so this right over here is four oxygens and then I have three water molecules each of them have one oxygen atom so three times one so this is going to be three right over here so on the entire right hand side I have seven oxygen atoms and on the left-hand side only have two so what can I do here can i what can I multiply by 2 to get to 7 2 times what is equal to 7 well 2 times 3 and 1/2 is equal to 7 so 2 times 3 and 1/2 is equal to 7 remember I have 2 here I'm saying 2 times something is equal to 7 I want to get to the 4 plus the 3 well I multiply it by 3.5 and now I have 7 7 oxygen atoms on both sides of my chemical equation but like we've seen in previous videos it is not standard to just leave a three and a half here it's kind of this weird notion of three and a half molecules we like to have whole numbers here so how do we how do we change how do we make sure we have all whole number coefficients in front of our molecules well we could just multiply everything by two then this thing is going to become a seven this thing is going to become a two this is going to become a four this is going to become a six so let's just do that I'll write the whole reaction over again so I have my ethane and I won't write actually what state it's in just just to save some time plus some molecular oxygen they combust the yield so these are the reactants the products are carbon dioxide gas and liquid water and liquid water so let's see you multiply if you say that there was a one out here before we're going to multiply that by two to get two we a three point five year multiply that by two you get a seven here we had a two right over here multiply that by two you get a four once again I'm just multiplying boats at all the coefficients by two just like you did in an algebraic equation in your algebra classes and then finally three times two is six and we're all balanced we were balanced here but we had we didn't have whole number of coefficients multiplying everything by two gave us the whole number of coefficients and we are much happier