 quite well. Obviously, probably the time limit was a little different this time. But probably looking at the scores, they gave you guys too much time, but it's no big deal. We can adjust that again. Anyways, let's go ahead and talk about more types of chemical reactions. So this time, if you guys don't mind, we'll talk about the type of reaction that's exclusively a redox reaction. So we call these single replacement reactions. They always occur when an element reacts with a compound. So you're always going to find an atom in its elemental state or a molecule in its elemental state by reacting with a compound. So you can see copper here is in its elemental state reacting with the compound silver nitrate there. And what you'll always find is in these redox single replacement reactions, the associated with the ionic compound gains electrons and stops being an ion, goes back to its elemental state and the element goes and replaces it and the ionic compound there. So you can see that hopefully. And hopefully you can tell that this is a redox reaction. Let's go over that and hopefully it's not too painful to figure out what's going on here. Hopefully, again, you guys are studying all the time. So we've got copper oxidation state of copper here. You guys remember? Zero. What about silver here? What is the oxidation state? Zero. How did you guys figure that out? They're in their elemental state. So those are zero and zero. So what you want to do, remember, is go from reactants to copper to something and from something to zero. Okay, so we know that already. Oxidation state of silver in this compound here. Does anybody know? What's oxidation state of silver, AG, in this compound here? Plus one. How did you figure that out? The nitrate is minus one, right? So silver here is going to be plus one. And copper here, what's its oxidation state? Plus two, right? Okay. Why does it have to be plus? Because it's a metal, right? You know that already. Not to mention that nitrate is the anion. Okay, so what got reduced and what got oxidized? You guys figure that out? So reduced means what? Gained electrons, right? Which one of these gained electrons? Silver. So this got reduced. Is the oxidizing agent or the reducing agent? Oxidizing agent. Oxidizing agent, yeah, very good guys. And you could probably do the rest, right? Oxidized. You should be able to do this with every reaction, of course. What I'd like you to look back again and look for all of those little cues as to tell you that this is a single replacement reaction. So just remember the things I told you. You got all this going to have an atom in its elemental state reacting with the compound. And I'm going to another atom in its elemental state and replacing with the first one where in the ionic compound, the cation, the region. So here's an example of, like a picture example, of course, of a single replacement reaction. This one's a little weirder. Just a hydrogen is being replaced, or the copper is being replaced by one of the, or both of the hydrogen. So a double replacement reaction is always going to be a non-redox reaction. You really find these between two ionic compounds that kind of do the dosy-doe and switch partners. So two compounds undergo an exchange of partners to produce two new compounds. So you see we go AB plus CD goes to AB plus CD. And you see that B and C are now partners and A and D are now partners. So B or A and C kind of switch partners. You can see these are all going to be examples of ionic compounds, and they've all switched partners, lead, nitrate, sodium chloride. If you go over these reactions, what you'll find is if you try to do the oxidation numbers of all of these different atoms in these compounds, you'll find that the oxidation numbers don't change. Therefore, it's an example of a non-redox reaction. It's always going to be the case. And here, hopefully, you can see the switching of the partners, Na for H. So Na goes to F, H goes to OH. This is also known as an acid-base reaction. Some of these double-displacement reactions are also known as acid-base reactions. So this top one up here is also an acid-base reaction. We'll talk more about those later. You can always figure out what if you've got an acid-base reaction because you're always going to get a salt and water reaction. So what I'd like you to do on your own, and maybe you've already looked at these on your own, is to classify each of these as a redox or non-redox, and buy its reaction type. Of course, if you're a redox, you can have three different types of reactions. If you're a non-redox, you can have three different types. I've written down what they are and the type of reaction. So I'd like you to go through those on your own, maybe looking at the different oxidation states, and trying to figure out that you can get those same answers. And then there's a variety of these that you can do on your own without answers. I would suggest you do them because I'm sure there'll be test questions on this type of stuff. And if you've got any questions, you can ask me during office hours. Okay, so let's focus more in on these double-replacement reactions or the double-displacement non-redox reactions. Okay, so these types of reactions are really kind of special types of reactions. Like I said, there's the acid-base type of double-displacement reactions. There's also this kind of precipitation reaction. And for those of you who are in my lab on Monday, we're going to go over that stuff that I was telling you about that we hadn't gone over, that we were going to go over at this point. Okay, essentially what we were going to go over is the precipitation reaction. So what is a precipitation reaction? It's a reaction in which two soluble compounds, so soluble, means able to be fully dissolved into the water. So you can see here, lead nitrate is aqueous, right? So that means it's fully dissolved into water. Sodium iodide is also aqueous. It's fully dissolved into water. So when you put these things into water, they break apart into their ions. When you mix these two solutions together, like you see up here, this is a clear solution prior to the solutions being poured together. And this is a clear solution up here, right? So both of these are clear solutions, but when you mix them together, what happens? We get this yellow stuff forming. That yellow stuff is the lead iodide solid here. And you also get the two other ions, the nitrate and the sodium, kind of coming together in a solution. But what's really interesting and important is the lead iodide. Notice that it's a solid, right? So we see it as it kind of doesn't really look like a solid there because it's just cloud of puff right now. But when it settles down, you definitely notice that it's a solid. So whenever you've got two solutions that are fully dissolved, that have fully dissolved ions in them and you mix them together and they form a solid, this is called a precipitation reaction. So when two soluble compounds react to form an insoluble compound, that insoluble compound we call a precipitate. So what's happening is the solvated ions are actually being removed from solutions. This will always be, or usually I guess it is, a double displacement reaction. I know it will always be a double displacement reaction. So what's happening here is that the tendency of the ions that become the solid have stronger electrostatic interactions for each other than they do for the water actually pulling them apart. So you know the reason why this thing is fully soluble in water is because water can easily pull it apart. So the attraction for lead to nitrate isn't very strong. But what you find is that the attraction for lead and iodine is much, much stronger. Just due to the relative chemical properties of these two ions, they're really attracted to each other. In fact, they're so attracted to each other that they overcome the ability of the water, the polar water molecules to rip them apart. And that's why they're forming this precipitate. So you'll usually see these double displacement reactions as either acid-base reactions or precipitation reactions. So unfortunately, for those of you who guys don't like to memorize things, this is something you're going to have to memorize because you've got to know when two ions come together which ones are going to form a precipitate. So this box of rules are something you're going to have to memorize. And it's really not all that bad. What you find is that most of the group 1 and group 2 ions are soluble. All the nitrates, acetates are soluble. A lot of halogens aren't soluble when they interact with lead, silver, copper, and mercury. And I think that's really all of them. So if you figure out which ones are insoluble, you're going to be able to figure out precipitation reactions. Because what I'm going to ask you to do is give me the state of matter. Maybe I'll give you a reaction equation without states of matter and you'll have to put in the states of matter. Or I could give you just the reactants and you'd have to give me the products. If you know this, you should be able to do that type of stuff. So let's go over a problem that predicts whether a precipitate will form or not. Let's just go over the problem that's on the slide. So we've got AGNO3, that's solid. That just means that then it's still in its reagent box. And KBR. So is KBR soluble or insoluble? Well we have to go look back at this chart. Well it says all common compounds of group 1A ions, lithium, sodium, potassium, and the ammonium ion are soluble. So is that a compound containing one of those things? Yes. So what you'll find is that it will always break up into its ions. Silver nitrate. So when we put silver nitrate into water, let's go over here and look at our rules. Nitrates are soluble. So does that pertain to this molecule here? Yes. Because it's got nitrate in it. So we know. But it's all different. You guys don't know the charges on your polyatomic ions yet. There's going to be a lot of that stuff on the next test. So you're going to have to go back and memorize your polyatomic ions and charges. So that's how we get the charge on silver is because we know the charge on nitrate. Okay so we know when we put these two guys into water, they're soluble. So we're going to have a beaker of this stuff, sodium nitrate, I mean silver nitrate, and a beaker of this stuff, potassium bromide. Then what we're going to do is mix these two together and ask ourselves does a precipitated curve. So what we need to do now, do we need to combine potassium and silver together? No. Because they don't combine together. They're two cations. So all we really need to worry about is the crossover combination. That and that. So let's ask ourselves, because when we mix these two, all those ions are coming into contact with each other. So we've got to ask ourselves, potassium nitrate, is this soluble or insoluble? How do you know that? I mean it says it up there. But potassium and nitrate are both soluble. So we know that this is soluble. Now let's look at the other one, silver bromide. If this is soluble or insoluble, because I was looking at the page, it's insoluble because it would have to be or the problem would be bunk. But if we look here, it says all common chloride, bromides, and iodides are soluble, uh-oh, except those of silver, lead, copper, and mercury. Is this one one of those? Yes, it's silver bromide, so it's insoluble. So if we look at all of these compounds, silver nitrate, potassium bromide, potassium nitrate, and silver bromide, the only one that would be insoluble would be silver bromide, so this would be our precipitate. And that's just a abbreviation for precipitate. Okay, so this is this. I would like you to go back and do this one on your own, the same kind of thing that we've done here. And then here's some more. Of course, these ones are a little harder because they say the names in the molecular form of the formulae. Okay, acid-base reactions. This is another form of a double-displacement reaction going on redox reactions. Combustion reactions and rusting reactions. These are both reactions with oxygen. You'll find that they're both redox reactions. Okay? I'd like you to go through those on your own trying redox numbers, okay, and proving to yourself that they're redox reactions. Okay, what I would like to do now is go over ionic equations. And we'll use this example that we've got on the board to talk about the various forms of reaction equations that you can write between a mixture of aqueous solutions. Okay, so for right now I'm going to erase most of this stuff. I'm going to go over the molecular equation, the total ionic equation, and the net ionic equation. Okay, so those are all things that we wanted to do for the... It should be able to do this. Okay, so what did we say? Was silver nitrate plus potassium bromide? That's the date. So remember, you've got to put that S there. To indicate that it's a solid. Plus, so this equation here that we've written down that shows all of the formula units, both reactants and products, this equation here is known as the molecular equation. So let's write the total ionic equation now. It's just all of the things that are soluble in water you break apart into their constituent ions. Okay? So out of those four compounds that are soluble in water that are on the molecular equation, how many are soluble in water? Three. Okay, so you're only going to break three of those up into their constituent ions. It doesn't break up into its ions, so we just say A, G, first wheel. So does everybody understand how we got the total ionic equation from the molecular equation? We don't. All we do is look on both sides and if something is the exact same on both sides then we cancel it out and don't write it in the ionic equation. So let's just go over here. So silver plus aqueous. Is that on both sides of this reaction equation? No. What about nitrate minus aqueous? Yes, so we just go. What about potassium plus aqueous? That's on both sides. What about bromine minus aqueous? No. Okay, so we're cool. Everything. All right, and then silver bromide solid obviously isn't going to be over here. Okay, so all we do is bring down the stuff that we didn't cross out. A, G plus aqueous plus Br minus ionic equation. The n-ionic equation always tells us the important stuff that's going on in the reaction, the actual reaction. The other things, these things that we crossed out, so what are they there? The potassium ionic and the nitrate crossed out. These are called spectator ions. They don't participate in the reaction. So does everybody understand what we've done here? Hopefully you guys can do it. I'm sure there'll be a test question on this stuff. Okay, so let's talk about this stuff now that we know how to do it. So does many reactions take place between compounds or elements that are dissolved in water? Just like how we have here. We've got two compounds dissolved in water. Ionic compounds, ladies, thank you. Ionic compounds in some polar covalent compounds break apart or dissociate into their ions just like what's happening here. Silver nitrate broke apart into its constituent ions or dissociated, the same thing, when they dissolve into water. So the equations for reactions that occur between these dissolved materials can be written in three ways. First, the molecular equation, which is the original equation up there. The total ionic equation, which shows the breakup of the original material into its constituent ions. And then the net ionic equation, which cancels out the spectator ions. Okay, so does that make sense? Hopefully it does. And here I'll let you go through it on your own. You can read about it. It just talks about, essentially, what we just did in kind of a more detailed format. And then there's an example given up here of the molecular total ionic and net ionic. Notice here, in this case, we've got my barium chloride, which of course has, which of course the formula unit is BACL2. So when we break that up into its ions, we're going to have to put 2 as the coefficient in front of chlorine, because there's 2 of them in the solution. So make sure you go over this and check that out. So amounts and chemical equations. The part everybody was waiting for, I'm sure. With chemical equations, the cool thing is, is that because the law of conservation of matter is obeyed, because we balance the equation, we show the relative numbers of each of the particular components of the equation. So, for example, when I write a hydrogen molecule, we need 2 oxygen molecules to make 2 water molecules. That's what we're actually seeing. That may seem obvious. Until you put it into terms of moles, and then everybody throws obvious out the window. So moles and particular molecules, you can think of kind of the exact same way. Mole is just a number, just like a dozen is a number, or a pair is a number. So if I have one mole of hydrogen gas, it's going to react with 2 moles of oxygen gas to produce 2 moles of water. Just like if I have one hydrogen atom molecule, it will react with 2 oxygen molecules to produce 2 water molecules. Just like if I have half a mole of hydrogen, it will react with 1 mole of oxygen to produce 1 mole of water. So, that's what the chemical equation tells us. Actually, it's the relative number of moles that's going to be reacting with each other and forming the particular product. So if I were to give you the mass of hydrogen, you should be able to give me the number of moles of water that was produced. So let's figure this out. So if I have, I don't know, what's somebody's favorite number? 121.7 grams of hydrogen. Mass of hydrogen is 121.7 grams. Or what's the mass of water that all makes? How do I figure this out? So let's figure out what's the mass of water. How would I do this? Do I just, well, so since this is a 1 to 2 ratio, do I just multiply this number by 2 up here? Can I do that? You've got to convert it to moles, right? Because they're not the same weight, right? H2 and H2O don't weigh the same amount, okay? But if I say a dozen of these and a dozen of these, that is the same amount. They're both 12 units, right? So that's what we're going to have to do is convert it to moles, which of course is in a dozen, but 6.022 times 10 to the 23rd, right? So the first thing we need to do is convert this number to a number of moles of hydrogen. How will we do that? How will we take the mass and convert it to a number of moles of hydrogen? So let's start there and then do what? Put grams on the bottom. Put grams on the bottom. And what else? One mole on the top. The molar mass of what? H2, which is what? Well, 2.016, right? So that would give me the number of moles of hydrogen, right? But remember, I don't care about the number of moles of hydrogen. I want to know the mass of water, okay? So let's figure out, well, how will we figure out the mass of water from that? Can we get the mass of water directly? No, but we do know this convergent factor here. Remember, one mole of H2 equals 2 moles H2O. Okay, that's given from the reaction equation here. The factor that we got. H2, 2 moles of water, moles of water, got it? But we don't want moles of water. We want a mass of water. So how will we get from moles of water to mass of water? Somebody else. How do I get from moles of water to mass of water? Yeah, so this is the same thing we did backwards before, right? So how will we do that? What will we be looking for on top here? Grams of what? Grams of water, of course, right? And what about the bottom? Moles of water, right? Specifically one mole of water, right? You guys don't know this. You're right behind, okay? Because obviously there are a bunch of people that do know this. What will we put up here? 18. And what is that? The molar mass of water, right? Yeah, 18.02. Water cancels out. And what does that give us for our units at the end? Grams of water, right? And is that a mass unit for water? Yeah, so that gives us the right amount that we want. So that's all we've got to do. Let's take 121.7, multiply it by 2.016, multiply that by 2, multiply that by 18.02, and that gives us... So remember, the chemical equation gives you this conversion factor that you need to use. For every one mole of hydrogen you have, you're going to make two moles of water, okay? So you could have done this same equation or the same problem and asked, well, what's the mass of oxygen? It's going to be a very similar process to doing what we've done up here. You're just going to take this, convert it to moles of hydrogen, but instead of converting that to moles of water, you're going to convert that to moles of oxygen, and from there convert that to grams of oxygen, okay? So it's just the same things you've been doing before that we did in Chapter 3 with all of this stuff. It's just we're using the chemical equation to get this conversion factor here, okay? So don't think of it as anything that's like totally foreign or totally new like we're learning a new concept. It's just the same concept just sticking this one step. So what I'd like you to do, here's that same kind of problem. There you go. You can do some more on your own. If you look here, I've got a series of calculations for the exact same problem. So what I've done here is I said, okay, we have 15 moles of carbon dioxide. How many moles of each of the other products and reactions do we have? And how do we go about getting that number, okay? And then if we look over here, well, if we have that many moles of each of these reactants then, what's the mass of it, okay? I think all of these should have a decimal point after there. So watch your sequence. I didn't put that decimal point there. And here's some more that you can try on your own. This reaction up here is a combustion reaction. So that's what this is talking about. Remember combustion is just taking anything that's an organic material so remember anything that has carbon in it is organic and setting it on fire, okay? In excess oxygen. So like you combust, I don't know, wood in a bonfire or whatever. In this case, this is your combustion of propane. So this is like a propane grill or something. And then here's another sample calculation except this time it's kind of like the two equations that I gave you on the last quiz yesterday. It's two elements forming an ionic compound, a metal and an ionic. Notice this equation is in balance either. So you're going to have to balance it, put the states of matter. So if ever I give you an equation, these are the things you've got to do is make sure it's balanced, make sure you put the states of matter in it and be able to go from grams and back and forth, okay? And of course be able to do a molecular net and total ionic equations. That's stuff. Oh no, no bad. Before we leave today, let's talk of, let's introduce the limiting reactant and we'll probably get through this. I know there's only like three slides left but we'll probably get through this next time. I just really want to introduce this. So the limiting reactant is the reactant and a reaction mixture that will run out first, okay? So what does that mean? That doesn't mean, okay, so let's look at this equation here again. Let's ask you what's the limiting reactant here? You couldn't tell me because you wouldn't know which reactant would run out first. Even though you might say hydrogen must be because it's only got a coefficient of one whereas oxygen has a coefficient of two. Well, that's not the way you look at this, okay? So limiting reactant is like, okay, let's pretend one of these things was more expensive, okay? So I want to use all of that up. I would want to just use a little bit of that and use it all up. So the other thing is less expensive so I can put a lot of it into my reaction, my reaction flask, to make sure all of the expensive stuff reacts, okay? So maybe if I have, let's say, let's pretend, I don't know, hydrogen is more expensive than oxygen, okay? So let's say I had one mole of hydrogen or I said four moles of hydrogen. How many moles of oxygen would I need to react with the four moles of hydrogen to make all of that one? Eight moles, okay? But remember, hydrogen is expensive so I want more than eight moles of oxygen. So let's add, but, ten moles of oxygen into my thing, okay? Would all ten of those moles react to form water? Not much, right? Because you only need two to one, right? So we would still have how many moles of oxygen left over? Two, okay? Two moles of oxygen left over. So you do this a lot of times when you're working with two things that one of them is more expensive than the other, okay? So what we say is that this would all react with this amount of oxygen, right? That makes sense, this would all react but we would still have this left over. So if we got this, we would form eight moles of this, right? But we would still have two moles of oxygen left over, right? Does that make sense to everybody? That makes sense, right? Okay, so what we would say is that the thing that's going to run out first in this case is what? Which thing is going to run out first? The hydrogen or the oxygen? Clearly, because there's still some left over oxygen over here, right? So we call the thing that's going to run out first, reactant. I usually call it the reagent, the limiting reagent. Reagent, reactant are almost the same thing but I will almost invariably call it the reagent. So let's just learn that so I don't have to keep thinking, oh, I mean reactant, okay? So we'll stop there. We'll finish up the last piece of this next time. I pushed that out back but that doesn't mean the test is going to be pushed back, okay? So the test is still next Friday so just start getting prepared for that especially with this reaction equation stuff. I'll leave a lot of test stuff on this, okay? So good luck guys.