Write a balanced chemical equation for the combustion of each of the following compounds:
Hey everyone in this question we're going to be writing out the combustion reaction for our decane. Now, maybe thinking how are we going to write out our combustion reaction if all they gave us was that our starting material or our hydrocarbon was decane? Well, for starters we know that decane has any molecular formula. Now, that is going to represent 10 carbons and it's a regular alkane, so we know that for our alkanes you can say the general formula for them and CNH to the 2n plus 2. So, all we need to do to figure out our molecular formula it is to know the number of carbons, which we said is 10, okay? So we're here we're going to say C to the 10, we have H to the 2 times 10 plus 2. So, how many hydrogen's do we have? 22, right? So let's write out our formula C10H22, okay? And in our combustion reaction there's many certain ratios of all of our different atoms, alright? Now in combustion all you need to know is that we're going to have a hydro carbon plus oxygen, okay? And now I'll probably take me back to your general chemistry days and you know what the products are for a combustion reaction? they're going to be carbon dioxide and water, so we're going to write a blank for carbon dioxide, which is CO2 and of course water, you know the formula for that, it's just going to be H2O and let's just move this over a little bit. Alright, so there we go. Now, all we need to do is simply write in water, alright? So, hopefully, this hopefully this is ringing a bell.
Now what we got to do is make sure we have this thing that we're to atoms on both sides, that means the same number of carbons on both sides, the same number of hydrogens and the same number of oxygens for both sides of our equation, right? So the same thing your C, H and O. Now, the easiest way to do this is to start with our hydrocarbon, we know it has how many carbons? 10, so we go over to the other side, where do we see carbon? right here, so we fill in the number 10, okay? Now, go ahead and do the same exact thing for your hydrogens and let me know what you get, we have 22 hydrogens on the left, so that means you need to have 22 hydrogens on the right and note that we have a H2 here, so the number that we write in is not 22 anymore, we have to write in 11, right? So that when we multiply 11 times that 2, we get 22, okay? So, so far we have 10 hydrogens, or excuse me, 10 carbons and 22 hydrogens on the left, 10 carbons and 22 hydrogens on the right, but now let's take a look at our oxygens, because in the process of writing numbers 10 and 11 that's going to affect our oxygen as well, right? Because this 10 going to apply only to our carbon but to our oxygen, so we expect 10 times 2, which is 20 oxygens for CO2 and 11 oxygen. So, 20 plus 11, we're going to expect 31 oxygens, so that means we used to go down to the left and write in how many oxygens? Well, here's where it gets a little bit interesting because notice it's going to be O2, so if we write in 31 here we have to multiply that by 2 and things will get the crazy, the easiest way to solve this is to write 31 over 2, okay? And then we need to make all of these numbers a whole number, okay? And remember that there is an imaginary one right here, so what we're going to do in our last step simply multiply our whole equation by 2 to cancel out this whole number because if we multiply 32 over two times 31 over 2 times 2 over 1, our 1's will cancel, our 2's will cancel out and it will just be left with 31 here then if we multiply 2 times 1 will get 2. So, so far our numbers are 2 over here, 31 and then on the right you multiply 10 times 2, so we should expect 20 and then 2 times 11, so we should expect 22. Alright, so this will be our complete combustion reaction with all of our atoms balanced down, the first blank will be 2 for our hydrocarbon, 31 for oxygen 20 for our CO2 and then 22 for our water, alright? So, I hope this makes sense and let me know if you have any questions on how we came up with a complete combustion reaction for our molecule decane.