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# Stoichiometry: Limiting reagent

Stoichiometry problem where we find the limiting reagent and calculate grams of product formed. . Created by Sal Khan.

Video transcript

Let's say we have some ammonia gas. That's NH3 and it's a gas, that's why the g is in parentheses. And we combine that with some oxygen, molecular oxygen, it's also a gas. And that reaction produces some nitrogen monoxide. NO, monoxide, there's only one oxygen there. That's also a gas, it's also called nitric oxide, not to be confused with nitrous oxide. I'll write nitrous oxide, it's N2O, Anyway, I don't want to divert you too much, we have to focus on the problem. So this is nitric oxide or nitrogen monoxide. It's a pollutant, I think it comes out of some cigarette smokes. I think it's used in the body, as well. You could we could look it up on the internet. There's your oxide. Ammonia is an important fertilizer. You know all about oxygen. And this also yields some water. So, plus some H2O. And we're told that we're given 34 grams of ammonia. And we're given 32 grams of oxygen. This is going to be the oxygen molecule, O2. So the question is, how many grams of nitrogen monoxide, or the nitric oxide, are going to be produced? So how much of just the NO is going to be produced in grams? So this is a stoichiometry problem. And so the important thing first is to just make sure we have a balanced equation before we even start anything. And lo and behold, we don't have a balanced equation. Let's confirm that it's not balanced. Let's see, we have one nitrogen here, we have one nitrogen there. That looks balanced so far. And remember, the pattern is start with the complicated stuff, leave the single atom molecules for last. Because those you can fix at the end without missing anything else up. Hydrogen, we have three hydrogens on the left-hand side. On the right-hand side we have two hydrogens. So let's see, how can we have three hydrogens on the right-hand side? If we multiply this times one and a half, 1.5, now we have three hydrogens on this side. 1.5 times 2, we have three hydrogens on the right-hand side. Things are looking good. We have two oxygen on the left-hand side. How many oxygens do we have on the right-hand side? We have one oxygen here, and we have one oxygen in this molecule, but we have one and a half of the whole molecule. So we have one and a half oxygens and then we have one more oxygen. So we have two and a half oxygens on the right-hand side, we only have two in this molecule. So what do we have to do? How can we get two and a half oxygens here? Well if we multiply it by 5/4, or 1.25. 5/4 times 2 is 5/2, which is 2.5. So now we have 2.5 oxygens. 5/4 times 2 is 2.5. 1 plus 1.5 is also 2.5. Well, we're not balanced yet. We can't leave this equation with just these weird decimal numbers over there. So let's rewrite it. If we wanted to get rid of all of these, we can multiply the entire equation by 2. No, not 2. We have to multiply the entire equation by 4 to get rid of this 4 in the denominator. So we multiply the entire equation by 4, we have 4 molecules of ammonia. We can even think of it in terms of moles. Right now I'm thinking of individual molecules. But we could say we have 4 moles of ammonia. 4 times 6 times 10 to the 23 molecules of ammonia. Either way it all works out. Hopefully, you're starting to see the value of moles. Plus 5 molecules, I'll just think of it in terms of individual molecules for now, plus 5 molecules of molecular oxygen. We're just multiplying everything by 4, that's what we're doing. Yields 4 moles of nitrogen monoxide. So 4NO plus, we multiplied both sides by 4, so plus 6 waters. 6 H2O. So there you go, we got some good practice balancing equations. So let's go back to the original problem. We're given 34 grams of ammonia. So what we need to figure out is how many moles of ammonia were we given? So what's the atomic mass of ammonia? Not the atomic mass, what's the molecular mass of ammonia? We're dealing with nitrogen 14, so it has a mass number of 14. Hydrogen has a mass number of 1. So each nitrogen has a mass of 14. And then the hydrogens each have a mass of 1. Remember, hydrogen is kind of strange. It doesn't have neutrons, or at least in its most traditional form. So it has just a mass number of 1. It's just a proton and an electron if it's neutral. So this is 3 times 1. We have three hydrogen atoms. So the mass, the atomic mass, of one molecule of ammonia is 14 plus 3 is 17 atomic mass units. Or, another way to write that, if one molecule of ammonia is 17 atomic mass units, then 1 mole of ammonia is how many grams? It's going to be 17 grams. So how many moles of ammonia are we given? We're given 34 grams of ammonia, 1 mole is 17 grams. So we're given 2 moles. 34 is 2 times 17. So this is 2 moles. We're given 2 moles of ammonia. Let's see how much oxygen we're given. Or how much of the molecular oxygen we've been given in this case. So let's see, the mass of just oxygen by itself is 16. The mass of just the atomic oxygen, you have to be a little bit careful here because sometimes people say we have 32 grams of oxygen when they're really talking about the molecular oxygen. Well, I guess it doesn't matter either way. But, sometimes when they talk about oxygen you have to make sure whether it is molecular or atomic oxygen. But the atomic mass number of oxygen is 16. I can confirm that by looking at the periodic table down here. So what's the molecular mass of the diatomic molecule O2? Well it has 2 oxygen, so it's going to be 2 times 16 equals 32 atomic mass units. One molecule of O2 is 32 atomic mass units. Or 1 mole of O2 is how many grams? Well if one molecule is 32 atomic mass units, then 6 times 10 to the 23 of that molecule are going to be that many grams. 32 grams. So how many moles of oxygen have we been given? We've been given exactly 32 grams of oxygen, which is exactly 1 mole. So we've been given 34 grams of ammonia, which is 2 moles. Let me write that in a nice vibrant color. And we're given 2 moles of NH3. And we've been given 1 mole of the oxygen molecule. Now, when we look at this reaction, for every 4 moles of ammonia, we need 5 moles of oxygen. Or for every 5 moles of oxygen we need 4 moles of ammonia. So something doesn't gel here. Normally, we need more moles of oxygen than we have ammonia. In the example that we're working through, we've been given less moles of oxygen than ammonia. We've been given less oxygen than we need for all of this ammonia. In an ideal world, if we had 2 moles of ammonia, we would need 2.5 moles of oxygen. The ratio of ammonia to oxygen, let me write that down. In a different color, this is getting boring. The ratio of ammonia, NH3 oxygen in our balanced equation, let me put that in a square so you know this is the most important part of what I'm writing. The ratio in this reaction is 4:5. So if I'm given 2 moles of ammonia, this is equal to what? If I'm given 2 moles of ammonia, how many moles of oxygen do I need? I need 2.5 moles of oxygen. Whatever this is, is that right? Yeah, 4/5. 1.25. If you doubled both of these numbers, you get 4/5. So I need 2.5 moles. But I don't have 2.5 moles of oxygen. I only have 1 mole of oxygen. So oxygen is going to be the limiting reagent in this reaction. I don't have enough oxygen. I have plenty of ammonia, but I don't have enough oxygen to react with it. So this is the limiting reagent. And I said before, the word reagent and reactant are used interchangeably. But when people talk about the limiting ones, they tend to call it the reagent. So oxygen is the limiting reagent. So we have extra ammonia. So given that we have 1 mole of oxygen, how many moles of ammonia can I react with that? So this reaction is going to look something like this. I only have 1 mole of oxygen. So instead of 5O2, I have to write 1O2. Let me make sure that's not of 10. Let me do it in a different color. I only have 1O2 instead of 5O2. So how many ammonia are going to react with that? Well the ratio is 4:5. So I'm going to have 0.8 ammonia. 4 is to 5 as 0.8 is to 1. And so essentially, if I take this whole equation up here and divide it by 5, I'll get what's actually going to happen. So this divided by 5 is 0.8, nitric oxide or nitrogen monoxide. Plus 6/5 moles of H2O. And so the original question in the beginning is, how many grams of nitric oxide are we going to produce, or nitrogen monoxide? So we have 1 mole of oxygen, 0.8 moles of ammonia, and we're going to produce 0.8 moles of nitrogen monoxide. Because we only have one oxygen. So 0.8 moles of nitric oxide or nitrogen monoxide. So what's the atomic mass, the molecular mass, of nitrogen monoxide? Nitrogen has 14 atomic mass units. Oxygen is 16. We've done that before. But you can confirm, your oxygen is 16, nitrogen is 14. So one molecule of nitrogen monoxide is equal to 30 atomic mass units. 1 mole or 6.02 times 10 to the 23 molecules of nitrogen monoxide, therefore, will be 30 grams. And how many moles are we producing in this reaction? Because oxygen was the limiting reagent, we only had 1 mole of oxygen here. Because of that, we can only produce 0.8 moles of nitrogen monoxide. So 0.8 moles of nitrogen monoxide, 1 mole is 30 grams. So 0.8 moles of NO is going to be equal to 0.8 times 30, which is equal to 24 grams. So we're going to be able to produce 24 grams of nitrogen monoxide. And so you might ask a question, we're only using 0.8 moles of ammonia. We were given, in the original problem, 2 moles of ammonia, so what happens with all the leftover ammonia? Assuming we mix it really good, we're going to literally end up with 1.2 moles of ammonia just doing nothing at the end. So we're going to have 24 grams of nitric oxide. And then we used 0.8 moles of ammonia. And we're going to have left 1.2 moles of ammonia. And if you want, you can figure out how many of the original grams of ammonia that is. You just figure out how many grams 1 mole of ammonia is, it's 17 grams. And then just multiply that times 1.2. Anyway, I hope you found that interesting. If you found this a little bit confusing, just watch the video again and pause it and try to solve it yourself. But I think it should start making a little bit of intuitive sense. I think the hard part is just the conversions between moles and grams and getting that part right. And then making sure that you understand the ratios. You understand this is a 4:5 ratio, there's always going to be less ammonia than oxygen. And if I only have 1 mole of oxygen, I'm going to have to have less than 1 mole of ammonia, even though I was given 2. So oxygen, in this example, was the rate limiting factor. Let's say I had 10 moles of oxygen and 2 moles of ammonia. In that case, ammonia would have been the rate limiting factor. Because then I would have had more than enough oxygen. Because for 2 moles of ammonia, I only need 2.5 moles of oxygen. So 10 moles would have been overkill. Anyway, see you in the next video.