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0:02
Commodore St. Germain: Hey, welcome back to another video of Chem with Com. Commodore St. Germain: Today we're going to be doing stoichiometry. This is a four-part series. This is the first one, and this is just called Stoichiometry Basics. Commodore St. Germain: Now, this is one of the most abstract and difficult concepts for chemistry students to, kind of. Commodore St. Germain: Absorb and really understand.
0:38
Commodore St. Germain: But at the same time, it's one of the most basic, basic math things that we'll do in chemistry, so it kind of is counterintuitive, right? But…
0:47
Commodore St. Germain: Give yourself some time to learn this, some space, some grace, right? Because you're gonna have difficulty having it click, but once it clicks, you're all good, so… just maybe watch this video a few times, take other notes, look at other videos, but I'm gonna try and break this down into a really understandable, Commodore St. Germain: format, alright? Commodore St. Germain: Alright, let's get started here.
1:15
Commodore St. Germain: Alright.
1:17
Commodore St. Germain: Stoichiometry Basics, again, 4-part video series, maybe do this one a Commodore St. Germain: I recommend not moving on until you really understand this, alright? Commodore St. Germain: Okay, so this is sort of a continuation from Topic 6, so go back and review that. It's measurements, prefixes, and dimensional analysis. It's the very last part. Commodore St. Germain: This is a continuation, sort of, of dimensional analysis. It just, it's very, very related.
1:45
Commodore St. Germain: And again, look at that Part 4. And this is what we left off on. This is a little conversion table I made. We're gonna build on this and make a more complex one, but for now, this is what we left off on on Topic 6, Part 4 of 4. So review that video if you need, some help with that. Commodore St. Germain: Okay. Commodore St. Germain: Stoichiometry expands on this table, or this conversion kind of tutorial I set up.
2:12
Commodore St. Germain: Using the mole ratio. Okay, we're adding this new number to our conversion table, and it's basically use the mole ratio of the chemical equation. So we're going to look at a chemical equation. I gave you an example here. Commodore St. Germain: And we're just looking at the ratio of the different materials, the different reactants and different products within that chemical equation. Commodore St. Germain: Right?
2:37
Commodore St. Germain: So, for example, the ratio for H2, diatomic hydrogen.
2:44
Commodore St. Germain: 2 ammonia, NH3 is a 3 hydrogen, because that's what we have here in the chemical equation, and 2 ammonia, right? 2NH3. So that's… that's all we're going to do. We're going to use those small ratios to be able to… Commodore St. Germain: Go back and forth between different products and reactants within a chemical equation. Commodore St. Germain: Alright, so let's go back to food, because this is what I like talking about. A recipe…
3:15
Commodore St. Germain: Lets you know how much you can make, or how much you need of things, and how much you can make of something. Commodore St. Germain: Right? That's what a recipe does. You know, the… for example, we have a recipe to build a sandwich. Commodore St. Germain: This sandwich… Is made by using two pieces of bread. Commodore St. Germain: and one piece of cheese. Reassembling those parts into another
3:46
Commodore St. Germain: another thing, which is the sandwich, right? And that's the ratios we're talking about. So we have 2 bread.
3:53
Commodore St. Germain: to one cheese, right? We know that. Commodore St. Germain: We also know there is one cheese for every one sandwich.
4:07
Commodore St. Germain: We also know there is 2 bread for every 1 sandwich. Commodore St. Germain: Right? These are the ratios of ingredients to other ingredients. Commodore St. Germain: of ingredients to products, and of other ingredients to products, right? You can go back and forth between all of these. What's really important is recognizing these,
4:30
Commodore St. Germain: numbers in red as what's called the stoichiometric coefficients. I know that's a lot of vocabulary, but these are the coefficients. We've talked about them before, especially when balancing. Commodore St. Germain: And they're gonna tell you, again, how much we have to use to make a certain amount, right? So let's take a due question. If we have 10 bread, right, if we have 10 pieces of bread.
4:56
Commodore St. Germain: How many pieces of cheese? Commodore St. Germain: do we need? We want to make a bunch of sandwiches, right? So if we have, 10 bread.
5:12
Commodore St. Germain: And we want to make sandwiches. What's the ratio of bread to cheese?
5:18
Commodore St. Germain: Well, we know that. Commodore St. Germain: The ratio is 2 bread to 1 cheese. Because bread is here, I want to put the bread on the bottom here.
5:30
Commodore St. Germain: 1… Cheese? Commodore St. Germain: Right, so we're comparing bread to cheese, because we know it's asking Commodore St. Germain: How much cheese do we need if we know we have 10 breads? So we're going from bread… Commodore St. Germain: to cheese. The bread is what we know. Commodore St. Germain: So, 10 bread, what cancels out? Bread cancels out, and you get 5… Jeez. Commodore St. Germain: Right? So we'll say 5.
5:59
Commodore St. Germain: Jeez. Commodore St. Germain: If we have 10 bread, we need 5 pieces of cheese to make all those sandwiches, right? Let's do another one.
6:15
Commodore St. Germain: Things are getting a little delayed here. Alright, let's do this one. Commodore St. Germain: If we have 20 sandwiches.
6:26
Commodore St. Germain: How many pieces of bread do we have? So here, we're going from… let me erase this part…
6:35
Commodore St. Germain: But this specific question, they're asking, hey, If we have 20 sandwiches.
6:43
Commodore St. Germain: how many pieces of bread did we need to make those 20 sandwiches? Well, we're gonna use the mole ratio of… Commodore St. Germain: Bread to sandwich, so this one. Commodore St. Germain: Because those are the two items that we're interested in. We know how many sandwiches we have, we're looking for how much bread we needed to make those sandwiches.
7:04
Commodore St. Germain: Alright, so we're gonna put the one sandwich On the bottom?
7:12
Commodore St. Germain: And the two bread. Commodore St. Germain: Right? Commodore St. Germain: Well, look what cancels out. Sandwiches cancel out, and it gives us our answer. We have 40
7:23
Commodore St. Germain: Pieces of bread.
7:29
Commodore St. Germain: Right? So we'll say… 40 breadth.
7:36
Commodore St. Germain: Alright, let's do the last one.
7:39
Commodore St. Germain: If we have… I mean, just erase this again.
7:52
Commodore St. Germain: Sorry, it's so slow today. Commodore St. Germain: Oops. Commodore St. Germain: Okay. Commodore St. Germain: Let's do this last one. If we have 3 cheese… 3… Cheese… Commodore St. Germain: How many sandwiches can we make if we had unlimited red?
8:13
Commodore St. Germain: Right? Commodore St. Germain: Okay, so we're gonna use now cheese to sandwiches.
8:20
Commodore St. Germain: Because we have 3 pieces of cheese, unlimited bread. We don't know how much bread we're gonna need for this, but we have unlimited. We're not gonna run out. We're gonna use that ratio. Since we have cheese on top here, we're gonna put cheese on the bottom.
8:36
Commodore St. Germain: Right? And you could probably have done this without doing, the math, but… Commodore St. Germain: it's very, very important to do these, easy examples so that when you do the difficult problems, you have somewhere to come back to, to kind of ground yourself, right? Because it gets pretty overwhelming here for this chapter. Like I said, it just…
9:01
Commodore St. Germain: It's not you guys learning, it's just the way chemistry is. It's a very, very difficult, vague topic, but with simple math, right? All we're doing is these small fractions. Alright, so we're going to cancel out the cheese, and then we can make 3… Commodore St. Germain: Sandwiches.
9:18
Commodore St. Germain: Right? And this is the slide you want to come back to, the video you want to come back to, when you get stuck with these more difficult concepts, right? Commodore St. Germain: And… we're finished there, right? So we calculated all that. Commodore St. Germain: Using what's called the ratio.
9:35
Commodore St. Germain: Right? Of ingredients to products, or even within the two ingredients, or it could be two products. It doesn't matter. It lets you relate everything in that recipe, right?
9:46
Commodore St. Germain: So a recipe lets you know how much you need of things, and how much you can make of something, right? That's basically what a recipe is. Now, if we go forward just a little more chemically, right, a balanced equation lets you know how much you need of things, and how much you can make of something. It's exactly the same thing.
10:05
Commodore St. Germain: Now, what's more difficult is the terminology, and you're saying silver instead of bread, and you're saying sulfur instead of cheese, and you're saying silver sulfide instead of sandwich. You're just changing the words, but everything is exactly the same, remember? We had two pieces of bread, one piece of cheese, and we had one sandwich.
10:26
Commodore St. Germain: atoms of silver, One atom of sulfur, and you can make one molecule of silver sulfide. Commodore St. Germain: Right? And these are the ratios, right? We have 2 silver to 1 sulfur. We have 1 sulfur to 1 silver sulfide. We have 2 silver to 1 silver sulfide, right? And again, this is just like the sandwich
10:54
Commodore St. Germain: Example. This is a specific ratio called a mole ratio. We've talked about it before, but this is where we're really going to hammer it home. This is called the mole ratio, and this is the center of stoichiometry, right? Commodore St. Germain: Let's answer the exact same questions we answered earlier with just these atoms and molecules now. Alright, if we had 10 silver.
11:22
Commodore St. Germain: how many sulfur do we need? So we're trying to figure out… we're gonna go from silver… Commodore St. Germain: So we're trying to figure out sulfur, and we're going to use this, what's called the mole ratio, because it involves the silver and the sulfur. Commodore St. Germain: And I'm gonna change this to chemical formulas and element symbols, just so we could,
11:46
Commodore St. Germain: get a little more technical here, right? So we have 10 silver. How many sulfur do we need? Well, since silver's here on top, we want to have 2 silver on the bottom, and I'm going to change the words a little bit, just so we can get used to this. Commodore St. Germain: And I'll explain that in a few slides of why we're doing this, but for now, let's just say 2 moles of silver.
12:13
Commodore St. Germain: Okay. The mole ratio is 2 silver to 1 sulfur.
12:24
Commodore St. Germain: Right? Oh, yeah. Commodore St. Germain: So, let's figure this out. Commodore St. Germain: Right? Look what cancels out. Commodore St. Germain: The silver cancel out. Commodore St. Germain: And… moles cancel out, right? So then we're left with 5… sulfur.
12:42
Commodore St. Germain: I'm just gonna put 5 here. Oops.
12:50
Commodore St. Germain: Okay, let's look up here. Commodore St. Germain: Now look, 5 cheese, right? Exactly the same calculation. Let's do another one. Commodore St. Germain: 20 silver sulfides, let me clean this up a little bit.
13:11
Commodore St. Germain: Hopefully that worked. Alright.
13:17
Commodore St. Germain: Alright, so let's try 20 silver sulfides. So, just like sandwiches, right? We had 27 sandwiches, but here's 20 silver sulfides. We're trying to see how many silvers we have in there.
13:32
Commodore St. Germain: So, we're gonna use the silver…
13:36
Commodore St. Germain: It's a silver-sulfide ratio. And again, whatever we're trying to cancel out, we're going to put on the bottom of the fraction. So 1AG… Commodore St. Germain: 2S, and you can put mole or not, but I'll… I'm gonna show you… we're talking about the mole ratio, right?
13:54
Commodore St. Germain: And because… yeah, I'll show, like I said. Let me just continue this. One… Mole. Commodore St. Germain: Silver. Commodore St. Germain: sulfide.
14:05
Commodore St. Germain: to 2? Commodore St. Germain: More… Silver. Commodore St. Germain: Well, look what cancels out. Silver sulfide cancels out, and the moles cancel out, and we get… Oops, sorry. Commodore St. Germain: 40… Silver Adams. Sorry it keeps doing that.
14:25
Commodore St. Germain: And I'm gonna write here 40. And again, that was our last answer as well. Commodore St. Germain: And let's do this third one here.
14:36
Commodore St. Germain: Let's erase this…
14:48
Commodore St. Germain: Don't know why it keeps doing that. Sorry. Commodore St. Germain: Okay, so let's do this last one. We have 3 sulfur, or they're asking how many silver sulfides you can make if you have unlimited silver. So we're using that equation now. So we have… again, we're starting with 3.
15:09
Commodore St. Germain: sulfur.
15:12
Commodore St. Germain: We're multiplying by the mole ratio, we're going to put the sulfur on the bottom, so it's 1 mole sulfur.
15:22
Commodore St. Germain: That's probably irritating for you guys, too. Commodore St. Germain: Sorry. Commodore St. Germain: I think maybe it's hitting my phone or something.
15:31
Commodore St. Germain: Alright, and… oh, sorry, I wrote that out, it's supposed to be a symbol, I forgot. Sulfur cancels out, moles cancel out, we're left with now 3… Commodore St. Germain: Silver. Commodore St. Germain: Sulfides. Commodore St. Germain: Alright? Just like we had 3 sandwiches in the last one, right? 3 sandwiches. Commodore St. Germain: So it works exactly the same for these chemicals, right? And the…
15:56
Commodore St. Germain: The problem is, when you click over from sandwich to silver sulfide, you lose a lot of that mobility and that, ability to be able to use mole ratios. Commodore St. Germain: Alright, let's do this again.
16:10
Commodore St. Germain: The mole ratio works for food. Let's check this out. Commodore St. Germain: Alright, here we're making egg noodles, right? So we have a pack of eggs, 3 packs of flour. Commodore St. Germain: And that'll give us 2 boxes of noodles, right? And again, it's a 1 to 3 ratio between Commodore St. Germain: Pack of eggs. Commodore St. Germain: To flour, 3 packs of flour.
16:37
Commodore St. Germain: It is a 1 to 2 ratio, where we have 1 pack of eggs to 2 boxes of noodles. Commodore St. Germain: Right? And we also have the ratio of this here, 3 packs of flour to 2 boxes of noodles.
17:01
Commodore St. Germain: Alright, so yeah, let's look at the ratios now.
17:05
Commodore St. Germain: And let's just… we're just centering on this one, because we don't have enough space to talk about all of them, but I want to recreate this with something more complicated, right? We have 3 packs of flour to 2 boxes of noodles. Commodore St. Germain: Well… Commodore St. Germain: Let's say we had… we needed a bunch of noodles, right? We can increase that to three dozen packs of flour, and that'll allow us to make 2 dozen boxes of noodle, right? The only
17:35
Commodore St. Germain: Thing that changed was this dozen.
17:39
Commodore St. Germain: Right? And you can… you need, you know, of course you need eggs with this, but this is just specifically talking about the ratio of flour to noodles. And you could do eggs to noodles, and you could do eggs to flour, right? Again, we could be very mobile with this. We just have to make sure we're very confident in what we're doing, and that's what we're doing right now, right? We can't just…
17:57
Commodore St. Germain: kind of putz around here, but we really need to put the numbers and the units and everything in the right spot. Alright, let's say, boom, we served it at a party, now Costco was, like, maybe the CEO of Costco was there, and he's like, you know what, I love these noodles, scale it up, we're gonna sell them in Costcos across the country. Commodore St. Germain: Right, so we increased that to now 300 packs of flour.
18:22
Commodore St. Germain: And we… that'll allow us to make 200 boxes of noodles, but the ratio is still the same. It's still a 3… Commodore St. Germain: to 2 ratio, right? That's the ratio. It doesn't matter what we put in front of that. Commodore St. Germain: And… Commodore St. Germain: we can say 3 moles of packs of flour, and 2 moles of boxes of noodles, right? That would be feeding the entire world
18:50
Commodore St. Germain: for the next billion years, right? There's so many in a mole. And this is how much a mole is, right?
18:56
Commodore St. Germain: So that's all you're saying, is when you're talking about the mole ratio, you're just talking about a huge amount, but that's called the mole ratio, right? And I just wanted to show you the value of a mole, which is 6.022 times 10 to 10 to the 23rd. Commodore St. Germain: anything. But again, this mole ratio works for food. And let's try it with, now… Commodore St. Germain: Chemical equations. We're going to do exactly the same thing we did.
19:26
Commodore St. Germain: And let's do that here. So we have 1 pack of eggs, 3 packs of flour, makes 2 boxes of noodles. Commodore St. Germain: Alright, let's see, video panel gone.
19:41
Commodore St. Germain: Alright, so this is the ratio, right? And we have 1… Commodore St. Germain: Diatomic nitrogen to 3 diatomic hydrogen. Commodore St. Germain: makes 2… ammonia, or NH3, right? And if you count the atoms, it's all there. We're starting off…
20:00
Commodore St. Germain: with 2 nitrogens, we're ending with 2 nitrogens. We're starting off with 6 hydrogens, we're ending with 6 hydrogens. But we're looking, again, at the products and reactants, and we're looking at the ratio of specifically those molecules, right? Commodore St. Germain: So here, we have 1 and 2 to 3H2. Commodore St. Germain: That's the mole ratio of reactants to each other. Commodore St. Germain: Here we have 1 diatomic nitrogen to 2 ammonia.
20:35
Commodore St. Germain: And here we have 3.
20:39
Commodore St. Germain: H2 to 2… NH3, right? That's the ratio.
20:46
Commodore St. Germain: Okay, and again, we're gonna expand on this last one, just because we don't have room to do all of them, but it works the same for all of them. Commodore St. Germain: So you have 3H2 to 2NH3. You also can do 3 dozen H2 to 2 dozen NH3. Commodore St. Germain: You can do 300H2 to 200 NH3. Commodore St. Germain: 3 moles of H2 to 2 moles of NH3. Again, this is all a 3 to 2 ratio.
21:24
Commodore St. Germain: And this is what we're going to be using, is the mole ratio for all of them. And again, this is just the number representation of a mole, just to keep you kind of focused on just a number of things, right? Commodore St. Germain: Okay, the mole ratio works for molecules as well. Commodore St. Germain: Right? This sandwich, in a sandwich, we know there are
21:47
Commodore St. Germain: 1, 2 pieces of bread, and 1 piece of cheese. So, mole ratio, 2 bread, one cheese. One cheese, one sandwich. 2 bread, one sandwich. Commodore St. Germain: Well, this works within a molecule as well. Let's look at water. Water has 1, 2 hydrogens, and one oxygen. So the ratio is 2 hydrogen to 1 oxygen.
22:15
Commodore St. Germain: one oxygen to… for each one water, there is one oxygen there, right? This is, for each one water, there are two hydrogens. Let's circle those. This is the oxygen. Commodore St. Germain: Right? And this is the oxygen. This is the two hydrogens, this is the two hydrogens, right? We're looking at ratios, and this is the mole ratio of a molecule, right? So it works for both.
22:40
Commodore St. Germain: So the mole ratio is used for calculations involving various products and reactants, as well as atoms within molecules. If you're looking for atoms within molecules, we've done this before already, back in chapter… I believe it was Topic 6, right? The dimensional analysis.
22:58
Commodore St. Germain: But we already… we've already used this. We didn't formally, you know, address it as long as this, because it was just kind of a simple concept, but this is where it gets a little more difficult. Commodore St. Germain: So, number 3 here is where you're gonna go from atoms within molecules, or molecules to atoms. You're going to use the mole ratio of the atoms within that specific molecule, like we did water here.
23:21
Commodore St. Germain: If you look at BAF2, you can see that there are two fluorines Right? Two atoms of fluorine. Commodore St. Germain: for every one molecule of BAF2, which is barium fluoride. Commodore St. Germain: Alright, we've done that before, just lightly, but now what we're going to do is we're going to introduce this other one here, where we're going to use the mole ratio, again, of the chemical reaction
23:47
Commodore St. Germain: to now be able to move back and forth between reactants and products, or between two reactants, or between two products, right? Just to figure out how much we need of stuff, how much we can make, and things like that. That's our…
24:04
Commodore St. Germain: Alright, so this is one important part when we're using this little table of conversions, right? We're gonna use this dotted line here as our separator between two different formulas, right? So let's say I want to look at, fluorine… Commodore St. Germain: Oops, let me fix that. Let's say I want to look at atoms of fluorine.
24:31
Commodore St. Germain: within one molecule of BAF2. Commodore St. Germain: And it says you're gonna assign one thing up top, and the second thing on the bottom, right? So here, I'm just gonna write flooring.
24:45
Commodore St. Germain: Here, I'm gonna write the whole BAF2. Commodore St. Germain: And since we're going from… Commodore St. Germain: atoms to within molecules, right? F is within barium fluoride, right? You can see it's in there. We're going to use the mole ratio of atoms within the molecule, so we're going to use this
25:05
Commodore St. Germain: Arrow here, and we're gonna go back and forth, or we can go back and forth, and that'll allow us to convert atoms to molecules, or molecules back to atoms, right? Commodore St. Germain: Alright, let's look at the other one now.
25:27
Commodore St. Germain: So let's say now we're… we have… Commodore St. Germain: We're trying to go back and forth between…
25:35
Commodore St. Germain: Let's say we have… we're trying to figure out… we have a certain amount of H2, and we're trying to see how much NH3 we can make, right? So let's say… Commodore St. Germain: H2 here. Since NH3 is a different formula.
25:51
Commodore St. Germain: you write it on the other side of this dashed line. Atom or molecule must have the exact same formula and cannot cross this dashed line without using number 3 or number 4, right? So we cannot cross that dotted line without using number 3 or number 4, and these are two different formulas, right? Commodore St. Germain: So, since they're two different formulas, you set them on opposite sides of this conversion table.
26:16
Commodore St. Germain: Okay, and we get through number 4 by using the mole ratio of the chemical equation. 3H2, which is here. Commodore St. Germain: 22NH3, which is here, right? So, again, we'll do some practice at this, later, but this is just showing you how to use this table and kind of giving you further instructions, right? Commodore St. Germain: Alright, let's continue. Commodore St. Germain: So here's a problem that's gonna come up, and it gets more complicated, right? So they're saying.
26:48
Commodore St. Germain: Title is, Moles of Reactant Required in a Reaction. They're saying, How many moles of I2 Commodore St. Germain: are required to react with 0.429 moles of aluminum. So they're saying, hey.
27:08
Commodore St. Germain: In this chemical equation, we know how much aluminum we have. Commodore St. Germain: We're trying to figure out how much diatomic iodine, or I2, we need to be able to react those four 0.429 moles. Let's look at the ratio here, right? Commodore St. Germain: 2 aluminum, 3 iodine, right? 2 to 3.
27:39
Commodore St. Germain: We could double that, right? But it's… we have to double both, and it's still a 2 to 3 ratio. Commodore St. Germain: You can multiply both by 10. Commodore St. Germain: It's still a 2 to 3 ratio. You could even go in decimals, right? You could have 1 here, but since it's a 2 to 3 ratio, you have to have 1.5 iodines, but it's still a 2 to 3 ratio.
28:07
Commodore St. Germain: Alright, you can do decimals here, smaller decimals.5. Well, it still has to be a 2 to 3 ratio, so it makes it .75 of iodine. Commodore St. Germain: Right? And here is where we're actually doing the problem, right? Commodore St. Germain: We're going to do .429 aluminum.
28:31
Commodore St. Germain: 2.6435 Commodore St. Germain: iodine, and that's still a 2 to 3 ratio, right? So how the problem works is, just like before, we get what we're trying to… we write what we're trying to do, or trying to convert, or trying to analyze, right? We're given Commodore St. Germain: They're saying, hey, we have .429 moles of aluminum.
28:59
Commodore St. Germain: Apply that by the mole ratio. Commodore St. Germain: And we're gonna put aluminum on the bottom.
29:08
Commodore St. Germain: Right? Because… and moles of aluminum cancel out with moles of aluminum, so we're left with now moles of iodine. Commodore St. Germain: So.429, let's break out a calculator here…
29:23
Commodore St. Germain: Hopefully you can still see this, so 0. Commodore St. Germain: 4 to 9? Commodore St. Germain: times 3… Divided by 2.
29:35
Commodore St. Germain: 0.6435, and that's what we got. Commodore St. Germain: 0.6435 moles of iodine. Commodore St. Germain: Right? And so what we did here is we went from aluminum.
29:50
Commodore St. Germain: to iodine. I2, right? This is our start, where we started.
29:57
Commodore St. Germain: Right? Because they're telling us, hey, it's 4.429 moles of aluminum. We're going to go to, now, moles of iodine. And we're going to use the mole ratio of the chemical equation, which was 2 aluminum to 3 iodine, and this is our end, and we're done.
30:19
Commodore St. Germain: Right? It's complicated, right? If you're getting confused here, go back a few sides, watch it over again. Stop it, make notes. Again, this is a very, very abstract concept. As you can see, the math isn't crazy difficult. It's 3… times 3 divided by 2, right? But conceptually, conceptually, this is a very difficult thing to have click. Commodore St. Germain: Especially right away. Commodore St. Germain: Took me a long time to get that to click for myself.
30:44
Commodore St. Germain: Alright, let's do some practice problems.
30:48
Commodore St. Germain: How many moles of H2 do we need to fully react to moles of N2? So let's write this down.
30:59
Commodore St. Germain: We have 2 moles of N2. Commodore St. Germain: We're trying to get to… moles of H2. Commodore St. Germain: So we're going from H2 to N2. Let's write down the ratios, and I'm going to write them in two different fraction forms. I'm going to say… Commodore St. Germain: 1… or let's do the equation first. 1 and 2. Commodore St. Germain: equals 3… H2. Commodore St. Germain: Right? This is what that means here. This 1, and there's no number, right? It's gonna be a 1.
31:34
Commodore St. Germain: I'm not sure if I can write up there, because it's too close to the top. There you go, 1. So the two fractions it's going to be are… Commodore St. Germain: We can either have 1 and 2… let's, I'm gonna… Fix this here. Commodore St. Germain: I'm gonna write down moles, because we want to be able to… Really… Commodore St. Germain: Focus on mole ratios. I want to make sure there's that word in everything. Commodore St. Germain: 1 mole and 2 equals 3 moles of H2.
32:12
Commodore St. Germain: Let's do two fractions. We're gonna say 1… mole… N2? Commodore St. Germain: over… 3 moles of H2. Commodore St. Germain: Here, we're going to say 3 moles of H2. Commodore St. Germain: over 1… mole of N2.
32:36
Commodore St. Germain: Alright, let's try it. Commodore St. Germain: So, we have… two… moles of N2.
32:45
Commodore St. Germain: We want to figure out H2, so we need N2 on the bottom, which is going to be… Commodore St. Germain: This one. Commodore St. Germain: So, we're gonna use that one. It's gonna be 3 moles H2. Commodore St. Germain: to 1 mole of N2.
33:04
Commodore St. Germain: Alright, so we're gonna look what cancels out. Moles of N2 cancel out. It's just 2 times 3 divided by 1 is gonna be 6 moles of H2.
33:23
Commodore St. Germain: Alright. Commodore St. Germain: That was number one. Let's do number two now.
33:29
Commodore St. Germain: Gotta figure out what's going on with that. Commodore St. Germain: I'm gonna draw a line to separate us. Commodore St. Germain: Alright. Commodore St. Germain: Number 2.
33:39
Commodore St. Germain: And it says, how many moles of H2 do you need to synthesize 12 moles of NH3?
33:52
Commodore St. Germain: That was a bomb here. NH3?
33:55
Commodore St. Germain: Maybe I'm pressing the button. Let's see here. Commodore St. Germain: And we're trying to get to… moles of H2.
34:06
Commodore St. Germain: All right. I'm gonna do this without making those fractions. I'm just gonna come up here and underline what I'm trying to use. So now we have 12… Commodore St. Germain: Moles… NH3.
34:22
Commodore St. Germain: I know, just because of units, I'm gonna need, on the bottom, moles of NH3.
34:33
Commodore St. Germain: And here we have a 2, so that has to be a 2. Commodore St. Germain: Okay, since I'm trying to get to moles of H2, I'm gonna need H2 up top, so I'm gonna say here, moles of H2.
34:50
Commodore St. Germain: And here I have a 3, right? It says 3, so we have to use the 3.
34:59
Commodore St. Germain: Alright, so look what cancels out. Moles of NH3. 12 times 3 is 36 divided by 2. I believe this is a, 18… Commodore St. Germain: Let's see, H2… what are we doing? NH3…
35:18
Commodore St. Germain: 12 times 3, 36 divided by 2 should be 18.
35:22
Commodore St. Germain: And that's gonna be moles of… H… 2.
35:30
Commodore St. Germain: That's how many moles of H2 we need to synthesize 12 moles of NH3. It had nothing to do with the other ingredient. We're just checking that one reactant, or the one ingredient, right?
35:42
Commodore St. Germain: When we do problems like this, we're just assuming we have unlimited of the other one, and we'll do more examples of that.
35:49
Commodore St. Germain: Okay, 3.
35:53
Commodore St. Germain: If we have 1.5 moles of N2, how many moles of NH3 can we make if we had unlimited H2? And that's very, very similar to problem 2. We just put that, Commodore St. Germain: last kind of phrase, can we make if we had unlimited H2? But it's the same thing. We're starting off with 1.5 moles of N2.
36:21
Commodore St. Germain: We're trying to figure out how many moles of NH3 we can make, right? We're just going from a different, we're saying,
36:37
Commodore St. Germain: Yeah, sorry, I think I misspoke over here. So, so, number two, right. Commodore St. Germain: Yeah, we're trying to synthesize 12 moles of NH3, but we're asking how many moles of H2 we needed. So this is almost the same thing, but now we're starting from an ingredient going to the product, right? Which is a little different, but it's the same exact concept. Commodore St. Germain: We're going to start with 1.5 moles of N2.
37:07
Commodore St. Germain: And, since we know… I'm gonna erase a couple things here, just so we could…
37:18
Commodore St. Germain: Show that, and on the bottom, just unit-wise, we know we're gonna have to have Commodore St. Germain: moles of N2. If you look at the ratio, since there's no number, that's gonna be a 1. Commodore St. Germain: We're trying to go to NH3, and here we are. 2 moles of NH3.
37:41
Commodore St. Germain: Alright, so… Let me let that to kick in. Sorry, it takes a little bit. Commodore St. Germain: A little bit cancels out. Moles of H… N… sorry, moles of N2 cancel out, and we can make… Commodore St. Germain: 3… moles of NH3.
38:04
Commodore St. Germain: Again, mole ratios, right? All you're doing is looking at coefficients. You're kind of… if you think back to our dimensional analysis and conversions, we're converting from one reactant or product to another reactant or product, right? Commodore St. Germain: And we're using R…
38:23
Commodore St. Germain: equation, our conversion rate, we're using our fractions, right? We're doing exactly the same thing. So if you want to think about this of… as converting one reactant to another reactant, or one reactant to a product, or product to a reactant, you could do it that way, and it's exactly the same procedure as converting. Commodore St. Germain: Right? And there's no difference.
38:46
Commodore St. Germain: The thing is, we had to make our conversion rates for that, right? So let's try another one. Commodore St. Germain: Alright. How many moles of calcium hydroxide are required to react with 1.36 moles of… Phosphoric acid. Commodore St. Germain: Right? To produce…
39:10
Commodore St. Germain: Phosphate, folateic, phosphoric acid, yep, and then to produce this calcium phosphate, right? So we're looking here…
39:21
Commodore St. Germain: 1… I'm gonna write the question down. We're trying… we're starting with 1.36… moles… of phosphoric acid.
39:34
Commodore St. Germain: And they're saying, hey, how many moles of calcium hydroxide do you need?
39:45
Commodore St. Germain: To. Make. Commodore St. Germain: this stuff, this calcium phosphate, right? And this is CAOH, close parentheses, 2, right? We're trying to figure out here. All we're gonna do is use the mole ratio, right? Commodore St. Germain: So we have 3 here. Commodore St. Germain: And we have 2 here.
40:07
Commodore St. Germain: So we're gonna start. 1.36… Moles H3PO4.
40:19
Commodore St. Germain: I guess maybe the wire's getting touched? Commodore St. Germain: I'll figure this out.
40:26
Commodore St. Germain: Alright, so now we're going to multiply by the mole ratio, and since we know this is on the bottom.
40:35
Commodore St. Germain: Oh, sorry, this is in the original value we're trying to kinda use to figure out the other value.
40:44
Commodore St. Germain: There is a 2 in front of that. Commodore St. Germain: phosphoric acid, so it's a 2 moles NH3.
40:53
Commodore St. Germain: And here is 3… moles… calcium hydroxide.
41:06
Commodore St. Germain: Alright, look what cancels out. This cancels out, moles of phosphoric acid cancel out. Let's do units first. Now we know we're going to be left with… Commodore St. Germain: Moles of calcium hydroxide. Commodore St. Germain: If you're doing your units, it's an easy way to check your work without having to actually check your work, right? All right, let's check the… let's do the math. So it's 1.36 times 3 divided by 2.
41:34
Commodore St. Germain: Let's use a different calculator. Commodore St. Germain: I think it's 1.36 times 3 divided by 2. Commodore St. Germain: But I got 2.04.
41:48
Commodore St. Germain: Let's talk about this real quick, right? If I have 1.36 moles of Phosphoric acid. Commodore St. Germain: I need 2.04 moles of calcium hydroxide. Commodore St. Germain: Right, and let's look at this. It's gonna be a… let's look at this, let's do it kind of…
42:13
Commodore St. Germain: Just visually, to see if we could figure out if this is actually true.
42:18
Commodore St. Germain: And the ratio is 2… Commodore St. Germain: H3PO4. Alright, let's do this. This is the ratio.
42:30
Commodore St. Germain: Let's do,
42:36
Commodore St. Germain: Alright, so we're starting with… Let's do this. Let's do, so it's a 2 to 3 ratio.
42:44
Commodore St. Germain: That's the same as, 4 to 6…
42:50
Commodore St. Germain: That's the same as a 1.
42:55
Commodore St. Germain: to 1.5? Commodore St. Germain: Right? Because it's… 2 thirds. And then our answer is 1.36. Commodore St. Germain: 2… 2.04. Commodore St. Germain: And these are all…
43:17
Commodore St. Germain: a 3 to 2 ratio. That's all we're really doing, is we're just figuring out the ratios between the reactants here. But you can go products, products to reactants, reactants to products. Commodore St. Germain: Alright, that's it.
43:31
Commodore St. Germain: Again, don't move on from this video until you understand at least 95% of this, because the other stuff, there's not much explanation, we're just going to do a bunch of practice problems, alright? So try that, and yeah, let me know if you need anything, but, I'll catch you next time. Thank you very much.
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