 So we're going to try a couple of problems that combine enthalpy with stoichiometry. The first one will be one involving enthalpy of solution and we're going to use an enthalpy value for a dissolving reaction, also called a dissolution, to work out how much energy is absorbed or released when something is dissolved in water. And in the second problem we'll look at an enthalpy of combustion and we'll use an enthalpy value and an energy value to work out how many moles of a fuel were burnt. And then we'll use stoichiometry to work out what mass of oxygen was needed to make this reaction happen. So let's try a problem to do with enthalpy of solution. I have a reaction equation here. On the left you can see we've got ammonium chloride solid and on the right of the equation I'm showing the two ions that the ammonium chloride is made up of, the ammonium and the chloride, and they're both shown with the subscript aqueous, meaning that the ammonium chloride solid has dissolved in water, the ions have separated, and we now have an aqueous solution of ammonium chloride. And you can see that that has an enthalpy associated with it. It's 14.78 kilojoules per mole. So if I dissolve one mole of ammonium chloride then the energy change will be 14.78 kilojoules positive. So that means that it's an endothermic process. And the question we've got is what will the energy change be if 45 grams of ammonium chloride is dissolved in water? So our strategy is this. The first thing we have to do, we have a mass of ammonium chloride and we have to convert that to moles. So we're going to calculate the moles of ammonium chloride that we have present. And the second thing is we will then use the relationship that enthalpy equals energy over moles to solve for energy because we have an enthalpy value and we will have a moles value. So let's try that out. So first of all, we know that we need to convert mass to moles. So we use our molar mass equals mass over moles relationship. And because we're looking for moles, we're going to rearrange that to make moles the subject of the equation. And we know then that we have 45 grams, so that will go on the top. And we need to calculate the molar mass of ammonium chloride. And you'll find that it is 53.5 grams per mole. And when we do that calculation, we find that that gives us 0.841 moles of ammonium chloride. Okay, now we know the moles of ammonium chloride. And we also have a value for the enthalpy of dissolution. So we can use the delta H equals delta E over moles relationship and rearrange it so that the change in energy delta E is the subject. And we find then that all we have to do is multiply the enthalpy 14.78 kilojoules per mole by the number of moles that we have present, which is a little under one mole, 0.841. And when we calculate that out, we find that it is 12.43 kilojoules. Now 12.43 kilojoules has four significant figures. We should look back at our calculations and find the value that has the lowest number of sig figs, that's the 45 grams, the mass of ammonium chloride. And we need to round that down to sig figs. So that gives us 12 kilojoules. We also should note that the original enthalpy value is positive. And that means that this energy change will be plus 12 kilojoules, which means that the system, the reaction is absorbing that much energy from the surroundings. So that means that it's an endothermic process.