 Now let's look at a different reaction, a combustion reaction. Here we're going to burn methane in oxygen to give carbon dioxide and water. The enthalpy of reaction this time is minus 890 kilojoules per mole and that means that if we run this reaction as written, that is with one mole of methane and two moles of oxygen, we will get 890 kilojoules of energy out. It's exothermic. That means that the enthalpy of combustion of methane has the same value. So remember, this was the enthalpy of reaction. The enthalpy of combustion of methane is essentially the same thing. We burnt one mole of methane and we got 890 kilojoules out. So delta H combustion of methane is minus 890 kilojoules per mole. Note that enthalpies of combustion are defined for one mole of a specific fuel. However, if we were instead interested in the energy as a function of the amount of oxygen that was used, we would have to do an adjustment. 890 kilojoules of energy is released for every two moles of oxygen used in this process. So the enthalpy value could also be expressed as delta HR, the enthalpy of reaction with respect to oxygen, would be 890 divided by 2. So that's minus 445 kilojoules per mole of oxygen for this reaction. We wouldn't call this the enthalpy of combustion of oxygen since the oxygen is not combusting. It is what is needed to make the fuel combust. It's what's called an oxidizing agent. More on this when you do redox reactions. Here's another combustion example, ethane gas burning in oxygen. Notice that this balanced reaction has two moles of ethane reacting with seven moles of oxygen. When we run this reaction with those amounts of reactants, we measure 3,120 kilojoules of energy being released. So what is the enthalpy of combustion of ethane? Well, two moles of ethane produces 3,120 kilojoules of energy when it reacts. So the enthalpy of combustion of ethane is 3,120 divided by 2. That is minus 1,560 kilojoules per mole of ethane. And if we wanted to know the enthalpy of reaction relative to the oxygen, we could easily calculate that also. So let's try a problem using that methane equation. We burn some methane and we measure that 12,600 kilojoules of energy are released. First of all, let's try and calculate how many moles of methane must have been burnt to produce that much energy. And secondly, what mass of oxygen would have to have been used in that reaction? So let's take part A first. This is a problem that's going to use the enthalpy relation. So enthalpy is energy over moles. And we already have values for the enthalpy. And we also have a value for the energy that's released. And what we're going to calculate is the number of moles of methane that must have been burnt. So we rearrange the relationship so that moles are the subject of the equation. And that gives us moles equals change in energy over change in enthalpy. We substitute in 12,600 kilojoules for the energy and minus 860 kilojoules for the enthalpy. And then we pause for a second. If we divide a positive number by a negative number, we're going to get a negative number. And I'm really not looking for a negative number of moles. That would be weird. So what have we done wrong? Well, the 12,600 value that we put in for the change in energy, it was specified in the question that 12,600 kilojoules of energy were released. That means it's exothermic. The energy is coming out of the system. So that means we need to put a negative sign in front of the 12,600. Now we have a negative divided by another negative. And we find that the answer is 14.16 moles of methane. We should round that off to 14 moles because our enthalpy value is to two significant figures. So we've got 14 moles of methane. We're burnt to produce that much energy. All right. In part B, we need to use what we already know to figure out how much oxygen was used. And this is a simple stoichiometry problem. So we know that the molar ratio of methane to oxygen from the equation is one to two. We know that we have 14.16 moles of oxygen. Remember that because I'm only part way through this particular calculation, I'm using the longer version of the number of moles. I'm not rounding off early. I will round off again when I get to the end of this particular calculation. So 14.16 moles of methane, we need twice as much oxygen. So that's 28.31 moles of oxygen. And then we simply use the mass-mol relationship to work out how much that weighs. So we get that the mass of oxygen equals the molar mass times the number of moles. You can work out for yourself that the molar mass of oxygen is 32.00. And we'll multiply that by 28.31. And that gives us 906.1 grams of oxygen.