 Alright, this is a little bit of a weird title, however this video, we are going to do one problem and probably it's the most difficult problem in the course. So that doesn't mean everything's going to be easy from here on out, but the problem that we're going to work on here requires you to pull in many different ideas that we've been talking about in the last bunch of videos. So let's talk about the hamburger again. So this time I go home to my wife and I say, honey, we're going to have a family reunion cookout. So I'm inviting all of your relatives, inviting all of my relatives, but don't worry about it. I got all the food, went to the store, I bought two hamburger patties, about twenty thousand slices of bread, we're going to cook them up, and we are going to have ten thousand hamburger patty sandwiches. Enough for you, me, your family, and my big Italian family. And what does she say? She starts yelling again, she starts saying, I should have married the other guy, who was a successful banker on Wall Street, yada, yada, yada, and I say, what's wrong? I got everything we need. And what does she say? She says, no, idiot, you did not get enough burgers, you did not get enough patties. You have plenty of bread, you probably have more bread than you'll ever need, but we can't make ten thousand sandwiches with the stuff that you have. And so I say, well, how many can we make? And she says, we can make two sandwiches. That's it, because we only have two patties. It doesn't matter how much bread you have. If you only have two patties, twenty thousand slices of bread, most of that's just going to go to waste. So what she has just described, and what I have just described, is called a limiting reagent problem in chemistry. The reason this is called a limiting reagent problem is because the number of hamburgers, number of patties, limits the number of sandwiches I can make. I can have a billion slices of bread, but since I only have so many, a few number of patties, this limits, how many sandwiches I can make, it limits it to two. And so maybe informally we would say that the patties are a limiting ingredient. The other term that we use here is this is in excess. We have more than we need to make use of all the patties that we have. This type of problem shows up in chemistry as well. Now I know that most of you could look at this and see it for what it is, that I didn't have enough patties, that I'm definitely not going to be able to make ten thousand sandwiches, and that I'm only going to be able to make two. I can make a maximum of two sandwiches because I only have two patties. However, I want to just briefly go over the more formal way of doing this. What you would do is you would break this equation down into pieces. You would say two patties needs four slices of bread. You would write it in reverse as well using this number. You would say twenty thousand slices of bread need ten thousand patties. If you look at the first question, we definitely have more than enough slices of bread. We need four slices for two patties, we have twenty thousand. So we can say bread is in excess. When we look at the second statement, twenty thousand slices of bread needs ten thousand patties. Well we only have two, so what we're saying more formally is two patties are limiting. What are they limiting? They're limiting our ability to make sandwiches. So because we found the limiting material limiting, and I know it was obvious we didn't need to do that fancy math to figure out which one we didn't have enough of, we can now use the limiting material to guess or to calculate how many sandwiches we can make. We can say two patties can make two sandwiches. And that just comes because the balanced equation would be one patty plus two slices makes one sandwich. And according to this balanced equation, one patty can make one sandwich, so two patties can make two sandwiches. So this long convoluted thing of looking at different amounts of ingredients being mixed together, we were able to sort of mathematically calculate that two patties, if we had twenty thousand slices of bread, still could only make two sandwiches. We are going to do that type of problem using a chemical equation and it's not going to be so pretty because I'm going to throw grams in there instead of these types of numbers which are equivalent to moles. So we're going to have to do a multigram conversion and we're going to have to find out which material is limiting. In this problem, it's pretty obvious. I made it ridiculously obvious which one was limiting. We have way excess amount of bread, the patties are limiting. In a lot of the chemical equation questions that are similar, it's not always going to be obvious which one is limiting and we have to resort to these sentences and this little bit of math to figure out which one is limiting. So let's go. So here's the equation. For the sake of time, I've balanced it already. So there's a four here, five, an arrow, four, and a six. If you wanted to read that out loud, you would say four moles of ammonia, ammonia is NH3, plus five moles of oxygen, which is O2, blah, blah, blah, blah, blah, blah, okay? So let's just keep moving on but you can think of those numbers as meaning moles again. I have two grams of NH3, 2.00 grams because somebody cared about significant digits and I have four grams, 4.00 grams of that stuff and I want to know how many grams of water I can make. Now what we have to do is we have to figure out which one of these we have extra of. Do we have, is this, is two grams of NH3 extra more than enough that we need to mix with four grams of O2 or is it the other way around? So first question we're addressing is which reactant NH3 or O2 is limiting? Now there's a lot of work that needs to be done. Because these numbers are written in grams and we want to know how many grams of water we can make, for me the easiest thing to do is convert all of these numbers, five moles of O2, four moles of NH3, et cetera, et cetera to grams. So let's do that step by step. We need to find the molar mass of NH3 because NH3 is our first molecule, molar mass of nitrogen and again we're just rounding to the nearest whole number to keep things simple. 14 grams for molar mass of nitrogen, there's only one nitrogen so 14 times one is 14 grams. Molar mass of hydrogen is one gram but we have three of them, so that's three grams, total of 17 grams of NH3 for every one mole of NH3. But we don't have one mole in our formula, we have four moles. So we have to, so four moles of NH3 weighs 17 grams times four which is 3468 grams. So again, that's an NH3, I'm writing the weights above the balanced equation, I'm making a new balanced equation using weights instead of moles. So let's keep going, we need the molar mass of O2 because we're dealing with O2, molar mass of O2 equals, well molar mass of O is 16 grams and we have two of them in O2 so that's 16 times 2, 32 grams of O2 per mole of O2. But we don't have one mole of O2 in our balanced equation, we have five so we have to multiply this by five to figure out how much five moles of O2 weighs, turns out that's 160 grams of O2. Next up, nitric oxide, molar mass of NO, molar mass of nitrogen is 14 grams, molar mass of oxygen is 16 grams, add them up 30 grams of NO per mole of NO. But we have four moles, we don't have one mole so if I want to know how much four moles of NO weighs, it's 30 times 4, that's 120 grams and then last but not least, molar mass of water, you'll probably do the molar mass of water so many times that you'll just memorize it but let's do it out, molar mass of H2O, molar mass of hydrogen is about one gram but we have two of them so it's one gram times two, two grams, molar mass of oxygen is about 16 grams, there's only one, so 16 grams total, 16 grams plus two grams is 18 grams so 18 grams of water is about one mole of water and but we don't have one mole of water in our equation, we have six moles so it's 18 times six, it's 108 grams so we have rewritten the equation instead of using moles, we've rewritten it as 68 grams of NH3 needs to mix with 160 grams of O2 and if we do that in those amounts, we can make 120 grams of NO and 108 grams of water, okay I've sort of cleaned up the slide and just rewritten everything here in grams so 68 grams of NH3 needs to mix with 160 grams of O2 and if you do that you can make 120 grams of NO and 108 grams of H2O so I want to talk about these two numbers for the moment and I want to write them as a sentence, 68 grams of NH3 needs 160 grams of O2, okay for the reaction to take place, those are the amounts that you need to use up both materials completely, however you can say, but I have 2.00 grams of NH3 in the problem, that's where that's coming from so 2.00 grams of NH3 needs, I don't know how many grams of O2 but because we know these two numbers we can figure out what this should be so we're going to resort to our favorite thing which is a ratio 68 grams NH3 needs 160 grams of O2 but we only have 2 grams of NH3 and so how many grams of O2 do we need so if we cross multiply, let's do it out, 68 grams NH3 times X grams O2, we multiply these two numbers together equals 160 grams O2 times 2 grams NH3, that's these two numbers multiplied together, we want to get the X all alone so we got to get rid of this guy, so we divide both sides by 68 grams NH3, divide this side by 68 grams NH3 and let's see on the left side they cancel, on the right side grams of NH3 those units cancel, the only units we're left with is grams of O2, grams of O2 on both sides, what we're trying to figure out is if I had two grams of NH3 like is stated in the problem, how many grams of O2 would I need to mix with it and that's going to be the answer to X, so it's going to be 160 times 2 divided by 68, turns out I would need about 4.71 grams of O2, I'm going to plug that in there just to make a sentence, so two grams of NH3 needs 4.72 grams of O2, but how many grams of O2 do I have in the problem? I only have 4 grams of O2, but if I want to use up all of this I need 4.72 grams, so what that means is this one, the O2, is the one that's limiting and this one over here, the two grams, this is the one that's in excess because according to this statement that we just figured out two grams of NH3 needs 4.7, needs to mix with 4.72 grams of O2, but the question says that you only have 4 grams, we don't have quite 4.72 grams of O2, we only have 4 because we don't have enough of this, this one limits, the O2 limits the amount of all of the other stuff that we can make on the right side of the equation, so we now know this is limiting, so now that we know the oxygen is limiting, the O2 is the limiting material, we can start writing sentences that relate this number to these numbers, now the question only asks about grams of water, so we can ignore how much NO we can make, but if we wanted to we could have figured it out, so I'm going to write a sentence relating this number to this number, what I'm basically going to say is 160 grams of O2 can make 108 grams of water, that just comes out from the balanced equation that we did, and this assumes, assuming we have extra NH3, but we already figured out that the NH3 was in excess, this was the stuff we had excess of, so this statement is correct, 160 grams of O2, if it's the limiting material can make 108 grams of water, and that's assuming we have extra NH3, which we did, we figured that out a few minutes ago, and so I can write this as a ratio, 160 grams of O2 can make 108 grams of H2O, but we don't have 160 grams of O2 in our question, we have four grams, so we can make an equal fraction off to the right, 4.00 grams of O2, how many grams of H2O can we make? Well, again we can make an estimate for what X is, 108 is less than 160, so X better be less than 4, so whatever our answer is, it should be somewhere around 1, 2, or 3 grams, so let's cross multiply to figure it out, this number times this number, 160 grams of O2 times X grams of H2O equals this number times this number, 108 grams of H2O times 4.00 grams of O2, we want to get the X alone, which means we need to get rid of the 160 grams of O2, divide both sides by 160 grams of O2, if we do that on the left side that reduces down to 1, on the right side grams of O2, grams of O2 cancel, the only units where I left with is grams of water, grams of water on the left and the right, and what is the question asking, how many grams of water can we make? So once we figure out this X, we will figure out how many grams of water we can make, so X is going to equal 108 grams of H2O times 4 divided by 160, so let's do that, and I come out with about 2, I come out with about 2.7 grams of water, so this was a lot of work to get to that number, we had to look at this equation, we had to convert the moles to grams and that took a while, so then we converted, we made a balanced equation using grams, we had to look at this and compare it to how much NH3 we had and how much O2 we had to figure out which one was the limiting material, by doing calculations, figuring out 2 grams of NH3 needs 4.7 grams of O2, we figured out we didn't have enough O2 and so the O2 was the limiting material, we didn't have enough of it, which means it's limiting, and we said, alright, so the O2, 4 grams of O2 limits how much water we can make, we said 160 grams of O2 could make 100 grams of water, well, if that's true, then 4 grams of O2 can make 2.7 grams of water, so that is a somewhat complicated problem, probably the most complicated one in the course, again, you know, there might be one or two questions like this on an exam, maybe one on the next exam and maybe one on the final exam, but that's about it, so I know that many students find this very difficult and very complicated, but that's it for limiting reagent, the rest of this unit is substantially easier than what we just covered. And I won't go to bed at all, I'll just lay there and wait, instead I'll make our favorite...