Balancing chemical equations
Current time:0:00Total duration:4:58
Balancing another combustion reaction
- All right now we have another combustion reaction. Instead of ethylene, we now have ethane, C2H6, has two carbons and six hydrogen atoms in each molecule of ethane, and it is reacting. It's ethane gas, it is 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 hydrogens. Here on the right hand side we only have two hydrogens. Here on the left hand side we have two oxygens. On the right hand side we have two plus, three oxygens, so none of the elements here are balanced. But like we did in the example of ethylene, whenever you see this where you have several 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 it having any other side effects on the number of carbons or hydrogens. So what I'm going to do is I'm going to first balance, like we've done before, the carbons and the hydrogens, which are going to have implications on the oxygens because if I change the number here, it's going to change the number of oxygens. If I change the number here it's going to 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 hydrogens on the left hand side, of the entire left hand side I only have six hydrogen atoms. On the right hand side I only have two right now, so 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 mauve color, right now on the left hand side I have two oxygens, but on the right hand side what do I have? Let's see, I have 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 I only have two. So what can I do here? What can multiply by two to get to seven? Two times what is equal to seven? Well two times three-and-a-half is equal to seven. So two times three-and-a-half is equal to seven. Remember I have two here, I'm saying two times something is equal to seven. I want to get to the four plus the three. Well I multiply it by three point five, and now I have seven, seven 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 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 to save some time, plus some molecular oxygen. They combust, they 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 had a three point five here, 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 all the coefficients by two just like you did in algebraic equations in your algebra classes. And then finally three times two is six. And we're all balanced. We were balanced here but we didn't have whole number coefficients. Multiplying everything by two gave us the whole number coefficients and we are much happier.