 Hey everybody, welcome to Tutor Terrific. Today I am going to do a chemistry video. It's kind of a three-part series on reactions and studying how to balance reactions, look at the different types of reactions. In this first video I'm just gonna do those two basic things and then we'll look at two particular types of reactions in more detail in two other videos. Today I want to show you the types of chemical reactions, the five basic types, and how to balance these five basic types of reactions. So here are the five types, combination, decomposition, combustion, single replacement, and double replacement. We're gonna look at those in that order. First, a combination reaction. We know we have a combination reaction when we have two elements and they're combined to make a compound. That is the basic type of combination reaction. You could have a compound over here or two compounds over there, which combine to make one larger compound over there. But the basic idea is you have more than one object over here and you have one object over here, a compound of some kind. So that's a combination. Take it two things and bringing them together. Decomposition then would be the exact opposite of that. We have a large compound or a compound on the reactant side and it breaks apart into its pieces. It's to its constituents. Maybe the cation, the positive part, and the anion, the negative part, are split. Okay, and that's a decomposition reaction. We have more objects over here than we do on the reactant side. Okay, and then combustion. This is a very specific type of reaction and there are a lot of options for it. We can have a hydrocarbon or some sort of organic compound that includes oxygens such as glucose, fructose, sucrose, any of the sugars or alcohols or any type of hydrocarbon that has some oxygens in it. And so that's the only variable. The other three, this O2 reactant and the carbon dioxide and water products are not optional. They are always present in a combustion reaction for these hydrocarbons. Okay, so this particular type of combustion reaction must have oxygen as one of the reactants and CO2 and H2O as the two products. And it must have some hydrocarbon or some organic compound with oxygen in it, C, H, and O in the reactants as well. Those are the type of combustion reactions that you could expect to see as a high school chemistry student. Then we move on to single replacement reactions. We're actually doing some swapping here, okay, in these last two. For single replacement, we have a single element which swaps with the cation or anion in another compound. And it can only do this if it's more active on this special chart. I'll show you called the activity series, then the other cation or the other anion, okay? And if it is, the reaction will occur and A will take the place usually of the cation, B, and this is what you have now, compound AC plus single element B. It took its place. In other words, it's like kicked it out for 10. This is like a couple and this guy wants to beat this guy out. And so they get in a fight and he wins and he takes the bow. That's one way to think of it. All right, now double replacement reactions. This involves two swaps that occur. Usually we think of the cations as switching anion partners. So we have two compounds, okay, and we end up with two different compounds. That's the idea here. That's how it's different than all the other ones. There are no single elements involved in a double replacement reaction. So we've got compound AB and compound CD. Now if the reaction occurs, it will yield, watch this, compound AD. Look, A switched partners with C and now A is now paired up with D. As you can see, plus CB, C is now paired up with B. Notice how I preserved the cation and anion order. The cations are listed first, even after the swap. So C, since it was a cation, was the positive part. It's still the positive part, even though it's got its different partner, B. And B and D are still listed second, even though in the original order, they might be reversed, such as B and C, for example. This is what trips a lot of students up. They think when A and D pair up, these two will pair up BC. But that's not true. B is a negatively charged part of the compound, and C is positive. So it would be listed first. Okay, now let's look at some particular examples with real elements and compounds. And let's bring in the balancing part of this video. All right, so here I have some reaction types for you. And I want you to help me determine what the type of reaction is and then balance it. Okay, so this first one, sulfur dioxide SO2 yields O2 plus SA. Now I made some of these up to make the balancing not trivial. And so don't quote me on the exact validity of these reactions. But as pretend it occurs, as pretend this reaction happens, and it exists, what type is it? Well, we have one compound over here, and we have two elements over here that came from the compound. So it's like they broke up into pieces. That would be a decomposition. D-comp, I'll just put that for short, decomposition. So that's what type of reaction we have. Now we have to balance it, okay? Now what that means for those of you who don't understand what I mean by that is we have to have the same number of each element on each of the two sides of the reaction, okay? Here's my method for doing this. You start on the left with the reactant. Start at the first and leftmost reactant and balance it first, okay? So S, I have one S on this side. No, this two does not count for the S. It counts only for the oxygen. So there's two oxygens and one sulfur on this side. If we look on the other side for sulfur, there's eight of them. And so this is not balanced in terms of sulfur yet. So what we have to do is we have to increase. We cannot subtract or decrease. We can only increase the side that has less. So sulfur on the reactant side has less. We need seven more, but we do not think of adding. We think of multiplying. So we need to multiply this by eight. So there's eight sulfurs on both side. Okay, now we look at oxygen. I know I've changed the number of oxygens because this eight, this coefficient eight that I added affects everything in the compound that follows it. So now there are two oxygens times eight, 16. There are 16 oxygens now on this side. I know sulfur is balanced, but now oxygen is not. There's 16 on this side and there's only two on this side. So I have to multiply this to get 16 because it has less right now than the reactant side. Obviously, that requires an eight. So now we double check again. There's eight sulfur on each side and 16 oxygens on each side. It is now balanced. If this reaction were to occur, it would occur with these ratios of the elements and compounds. All right, next one, potassium plus that's boron oxide. So di boron trioxide if boron was a non-metal and potassium oxide plus boron as an element. Okay, so this one, what do we have going on here? It seems like I have a single element on each side and a compound on each side, but was there any swapping that occurred? Yes, potassium seems to have switched places with boron. Boron is now on its own on the product side. So that means this is a single replacement reaction, single replacement, okay? The potassium has knocked boron off and now boron is by itself and potassium is bonded with the oxygen. It is however not balanced and so we need to do that, okay? Again, we're assuming all these reactions occur and not debating that for the purpose of this video. Okay, let's look at potassium first. Start on the left of the reactant side. There's one potassium over here, but there's two over here. So we'll come back here to where there's less and multiply that by the proper number to balance the number of potassiums. Two, okay. Now we move on to boron. Boron, there's two on this side, but there's only one on this side. So again, I'm gonna need to double the one that has less on the reactant side. Now boron is balanced, but oxygen, there are three oxygens here and there's only one oxygen here. This two only applies to the element directly preceding it, potassium as all subscripts do that do not use parentheses. We'll see some of that later. So, okay, we have to balance that oxygen now. So let's start by tripling this potassium oxide so that we have three on each side. Okay, great, oxygen's balanced. So they're all balanced, right? No, we're not done. The reason we're not done is because I just changed the number of potassiums on the product's side. Now there's six, three times two potassiums on the product's side. Now I have to go back to the reactant side and make six potassiums over here. So you can erase coefficients that you write and change them if you need to increase them later. So if I increase this to six, we should be good now. Six potassiums on each side, two borons on each side, three oxygens on each side. Okay, remember this three applies to everything in the compound after it. Everything's tripled. Okay, great, this one is finished. Next one, we have elemental iron plus chlorine, Cl2. I'll get more into why Cl has a two by it when it's by itself in another video. Iron plus chlorine gas gives us iron three chloride, okay? Iron three, why do I say iron three? Because chlorine has a charge of minus one when it's in ion form. That's its primary oxidation number. And iron, in order to balance with it, since there's only one of them, has to have a plus three charge. We're not going to do charge balancing in this video, but we are in a later video. All right, what type of reaction is this? Five iron and chlorine elements, and I combine them into a compound. I'm combining them into one thing. That's called a combination. So all right, CO and B for short. This is a combination, okay? So let's balance it. All right, we've got iron one over here. Let me come over here and look at iron. On the product side, there's one. So iron is already balanced. But if you look at chlorine, we got two chlorines over here and three chlorines over there. A two against three is not ideal, and it's going to require some work. What I recommend you do is not use decimals. Most teachers don't allow you to use a decimal coefficient. For example, some teachers would say, you could do 1.5 Cl2s, which would make three, and then you're done. But what most teachers require you to do is to use whole number coefficients only, natural numbers. So my recommendation to you is to double the side with chlorines that is odd. Cl3, that's an odd number of chlorines. We want an even number. It makes things easier. Now we have six chlorines over here, an even number. And then we'll come back over here since we just affected the number of irons, we finished balancing them. Now we have to redo it. There's two irons here. So we're going to need to put a two in front of this Fe to make two irons on both sides. Then chlorine, chlorine, chlorine, chlorine. There's six over here. What do I need to put in front of this chlorine to make six over here when there's two right now? A three, like so. So I have two irons on each side, and I have three times two, six chlorines on both sides as well. So now that's balanced. All right, three more examples coming up right now. All right, so this first example of my second set of examples, let's look at this specifically. We've got C2H4, which is actually ethane. And we've got, oh, excuse me, no, ethene. All right, and that's a much later video how to name these, C2H4. And then we've got plus O2, which yields CO2 and H2O. These last three compounds should remind you of a combustion. This is a combustion reaction. Those three, these three compounds or elements, oxygen, carbon dioxide, and water are always present in these positions in a combustion reaction for a hydrocarbon or some other organic compound with Cs, Hs, and O's. So this is a combustion. Now I'm not going to use the same abbreviation that I used for the combination. So I'll write out the whole word, combustion. Okay, so that's a combustion. Now these are famously the most challenging to balance when we have large hydrocarbons, but I'm trying to make this an introductory video, so we're going to use a simple one for this video. Now let's look specifically at the carbons first. There's two on this side, and there's one on this side. So the one that has one needs to be doubles to two. Okay, then we go, I know that affects oxygen, but we're going in order. Usually when we get really complex, like some of these down here, we save oxygen for last because it's present in so many of the compounds. This is a good example of when to do that. So we're going to do hydrogen next. There's four hydrogens over here, and how many are over here? Two. So we need to double this side, so there's now four on both. Now we can look at oxygen. Okay, so over here we have how many oxygens on the reactant side? We have two. How many do we have total on the product side? You have to be careful how you count. I have oxygen in this compound and this compound. In this compound, I have two oxygens here, but times two, that's four, then I have to add to that one times two oxygens. So two, which makes a total of six. Four plus two equals six. Okay, so there's six total over here, and there's only two over here. How can I multiply this to get six? I can multiply it by three. All right, double check everything. Two carbons, two carbons. Four hydrogens, four hydrogens. Six oxygens, four plus two, six oxygens on both sides. So this one's finished. All right, next one. We've got iron three bromide plus hydrogens sulfate. If it's dissolved in water, it's sulfuric acid. And we have iron sulfate and we have hydrogen bromide, which if it's dissolved in water is hydrobromic acid. Okay, what's happened here? Well, it looks like iron is now with the sulfate and hydrogen is now with the bromine after the reaction has occurred, which means there was a swapping of cations with anions. Have you ever seen the show wife swap? I think it was on TLC for a while. It might still be airing kind of a weird premise for a show, but it was entertaining. Two wives would swap husbands for a while and Lord it's like to live with someone who has different issues than what they're used to. Maybe they become more grateful for their husbands. Well, that's kind of what the idea is here. We've got two wives swapping husbands and that's what occurs here. Now there's different amounts. If you have a reaction that's already done, but not balanced because of charge balancing, I need three bromines, which are each minus one, for each iron, which is plus three. This is iron, Roman numeral three. Sulfates charge here and here is minus two. Since hydrogen is plus one, I need two of them to charge balance with sulfate. And over here, I've got a minus two against a plus three. And so I'll need to get to six, the least common multiple of both of these. And so we'll have iron sulfate in this ratio. And then hydrogen and bromine plus one, minus one charge. So they're a one to one balancing. Okay, now let's get to work on balancing this. There's two irons on the product side, but there's one iron on the reactant side. So the reactant side needs to be doubled, the whole compound that is. Then we check bromine. There are three bromines times two now, making six bromines. So we'll need, over here we have one. So we'll need to multiply this. By six, okay. Then we check hydrogen. Hydrogen. Oh, by the way, what did I say? This was a double replacement reaction. Double replacement because of the wife swapping. Okay. So six hydrogens. Over here I have two. So I'm going to have to multiply this by three to get six over here. Okay, obviously that affects quite a few things, but let's check. Three sulfurs now on the reactant side. On the product side, look closely. S3. So now we see a situation in which we needed three sulfate ions. These are polyatomic ions. And this parenthesis means that outside the parenthesis, that little number applies as a counting number for all of the elements inside. So for sulfur, there are three sulfurs. So sulfur is balanced. Now we need to check oxygen, the last one. So O4 times three means 12 oxygens total on the reactant side. On the product side, you see four inside the parenthesis and a three outside, which applies to the oxygen in multiplication fashion. So now we have four times three, 12 oxygens on the product side. So oxygen is actually balanced. So we check back with everything else. We're good to go. We didn't mess any balancing we did up earlier. So this double replacement reaction is now balanced. Now this last one. It's a little tricky. So let's look closely. We've got HCl, hydrogen chloride, or hydrochloric acid if it's in water, plus sodium OH, which is hydroxide, which makes sodium chloride plus water. Huh, okay. So did hydrogen and sodium swap anion partners? Well, the sodium now is definitely with the chlorine right here, sodium chloride. So that seems normal. But the hydrogen, we expect to see HOH over here. Well, water is essentially that. Although it's slightly different. Water can be made from a hydrogen ion and a hydroxide ion as well. So this is HOH, this H2O right here. So those are equivalent ways to write this even though this way it's a little trickier. So hydrogen did bond with the hydroxide ion to make water. So this is a wife swapping or a cat ion swapping double replacement reaction as well. Now let's check the balancing. Hydrogens, how many hydrogens over here? Two. How many hydrogens over here? Two. So I counted in H from each of the two compounds on the reactant side. Then chlorine. We got one chlorine over here. We got one chlorine over here. Okay, that's balanced too. This is quite strange. Then we have one sodium on the reactant side and one sodium on the product side. So that's good as well. And then lastly oxygen. I've got one oxygen here and I've got one oxygen here. Whoa. So this entire equation, this entire reaction was balanced already. We didn't do anything wrong guys. Sometimes that's how it is. Just by charge balancing which was already done the equation is balanced in full and we don't need any large coefficients like these other examples. All right guys, this has been a basic tutorial on bouncing and different types of reactions. In the next few videos we're going to look at single replacement reactions and double replacement reactions in particular because some complications do arise and some more things we can do with those reactions specifically. All right guys, thanks for watching this video. This is Falconator signing out.