 All right, let's talk about reaction types and how you use that to figure out what the products of a reaction are going to be to complete the chemical reaction, where I give you the reactants and you're going to tell me what the products are. First you're going to know what kind of reaction it is, and to do that you're going to look at what I give you. If I give you two elements it's going to be a synthesis reaction, and all you have to do is take those two elements, combine them together, and use the criss-cross method to write one formula for one new compound at the end. If I give you one compound, then it's decomposition, and you're going to break that compound down into two simpler substances. A lot of times it is just separating them back down into their elements again. You write down what the two elements are, put a plus sign between them, and then watch out for those diatomic elements. So like if it's chlorine you have to put Cl2. It's hydrogen you have to put H2. Any other element would just be its symbol. There are some special circumstances, some special conditions that I show you in class that I'm not entirely sure if you have to know for the quarter exam or the test itself. I haven't seen either one. So at this point on this sheet anyways all we're going to do is one compound breaking back down into two elements again. If I give you one element and one compound then that is going to be single replacement. And what you're going to do there is you're going to swap out like for like. So if I give you a metal as the element, you're going to swap it for the metal that's in the compound. You're going to swap places with that metal. If I give you a non-metal as the element, you're going to swap it out for the non-metal in the compound. That's what I mean by swapping out like with like. If I give you two compounds, then that's double. Then you just take the first half of the compounds and swap them. And you can do it either way. You can take the second half of both compounds and swap them. But the easiest way to do it is just to split the compound in half and swap the first parts and write your new formulas. And every time you're writing formulas on this it's going to end up being crisscross method because a lot of these combinations here, a lot of these reactions here are ionic. Almost all of these reactions here are ionic. So you'll use crisscross method to write your formula. Here you'd be writing the formula for one compound as your product, up here, synthesis. One compound as your product. Here you'd have to write what you see. One compound and one element as your product. Again, your compound's going to come from the crisscross. And in this one you'd be writing two compounds. And again, those two compounds would come from the crisscross method. You'd figure out the substitutes that way. The last type is combustion and it's going to be a hydrocarbon. See something H something plus O2? That's combustion. Those are the easiest of all of them to predict because all you're writing down is carbon dioxide and water. CO2 and H2O. Let me take a look at a few examples with you to kind of walk you through the process a little bit. We look for the pattern here. We try to see what we've got. See something H something plus O2? See something H something plus O2? That's a combustion reaction. And again in combustion, right down carbon dioxide, water, call it a day, move on to the next. Those are some of the easiest ones to predict. As long as you recognize that the hydrocarbon, the C something, H something, plus oxygen is a combustion reaction. Nothing really to figure out there, nothing really to do there. Just write it down. Let's take a look at this one. I have to analyze what I've got. Two capital letters there makes that a compound. Two capital letters there makes that a compound. Two compounds. Two compounds is a double replacement. And again what I'm going to do is I'm just going to split my formula in half and I'm going to swap these positive parts. So the H is coming here to combine with the Cl and the Cu is coming over here to combine with the S. Now I can't just write that down and say I'm done. I have to look at the subscripts of my formulas to figure out the charge of that copper. So I know what the crisscross there, it's a transition metal. So I had to figure out what the charge is over here with the copper chloride. So I know what charge to use when it combines with the sulfur. And then I got to use the crisscross method over here with the hydrogen and the chlorine. Hydrogen's in group one, it's a plus one. Chlorine's in group 17, it's a minus one. I got a couple of ones to swap out here. So it ends up being just HCl. So that one was okay. Now for this, the copper and the sulfur. Over on this side, copper is going to have some charge to it. I'm going to figure out what it is because that charge will stay the same when it goes over to the product side. Chlorine as we already said is a negative one. There are two of them for a total charge of negative two. We always start with our total negative charge. We know our total negative charge and our total positive charge has to be the same. So copper's side of the compound has to be a plus two. We divide it by the subscript on the copper and there was nothing there, so we assume that one is one. The oxidation number for the copper is a plus two. So this was copper two chloride. That two had been this Roman numeral in the middle. It's a plus two in this compound, which means it's going to be a plus two on the product side in the new compound as well. Sulfur, we just look up on the periodic table. It's in group 16. It's a negative two. We got a couple of twos here, which of course we reduced to ones. And that ends up being just CUS. So it was good to go also. So HCl plus CUS is the answer. So we'll move on to the next one. Again, we're trying to figure out what kind of reaction it is, so we know what to do with the elements here, how to rearrange them. Two capital letters, that means that's a compound. Three capital letters, that means that's a compound. So two compounds, compound plus a compound, two compounds is double replacement again. So once again, we're going to split them into positive and negative halves and we can swap out those positive halves. So the potassium's moving over here with the Cl and the barium is moving over here with the PO4. Again we just don't stop there. We have to crisscross. That's the number one mistake students make. They swap elements around just fine, but then they forget they have to crisscross to get the subscripts. So we look up potassium on the periodic table. It's in group one. That makes it a plus one. We look up chloride on the periodic table. It's in group 17. That makes it a minus one. Again we have a couple of ones to put in here and by rule we don't write down the ones. So it's just KCl. Over here in the second compound, BA, that's in group two. So it is a plus two. Phosphate's one of our polyatomic ions. Look it up on the chart. It's a negative three. So we have a two going over here and we have a three going over here. The BA three. That's simple enough. But remember before we can put that two on the phosphate, we have to put the phosphate in parentheses to protect it to make sure its formula doesn't change. So it's BA three PO four two. And again we write that down as our products. And we're done. We predicted it. Let's look at the next one. In the next one we have C something H something plus O two. C H something plus O two. That's combustion again. And whenever it's combustion, write down your carbon dioxide in your water, call it a day and move on to the next. Some of the easiest ones you can do. Again it's C something H something plus O two. That's how you know it's a combustion reaction. Let's look at the next one. Potassium and chlorine, two elements. Whenever I give you two elements it's a synthesis reaction. And all you're going to do is the crisscross method to figure out what the formula is. Just like we did up here. Potassium is a group one so it's a plus one. Chlorine is a group 17 so it's a minus one. We have a couple of ones to swap out. One there, one there. Don't write down the ones. It's just KCO. And that's it. You're done. It's a synthesis reaction. There's only supposed to be one product. Let's do one more. Let's do two more. Let's just sum down here to these two so that you know what you're supposed to be dealing with those. I only have one reactant here. One reactant is the key giveaway of a decomposition reaction. And again up here, if you only have one compound given to you it's a decomposition reaction. And in most of these all we're going to do is just split this up into its two elements. So it splits right down the middle. Sodium is the first element. It's just Na. Chlorine is the second element. It's one of those diatomics. And because it's diatomic you have to put the two on it. The diatomics are hydrogen, nitrogen, oxygen, fluorine, chlorine, bromine, and iodine. If they are by themselves in a chemical reaction you automatically put the two on it. If they're in a compound like they were up here we've got to do the crisscross method to figure out what to put on it. But when they're by themselves as an element you've got to put a two on it. And again I want to do this one as well so that you're familiar with what to do with this type of reaction. Two capital letters makes that a compound. One capital letter makes that an element. One compound, one element, single replacement reaction. So like replaces like. This is zinc. Zinc is a positive ion. It's a metal. So it has to replace the positive part of that compound like replaces like. So we're going to swap out the zinc and the hydrogen. So it's going to be zinc and chlorine as our new compound and hydrogen is going to be by itself. So to figure out the zinc thing, zinc is one of the rare predictable transition metals. It always has an oxidation number of plus two. Hydrogen's a minus one, two goes there, one goes there, ZnCl2 is going to be the chemical formula. So you write that. Hydrogen's going to be by itself but hydrogen is one of the diatomics. You have to put the subscript two on it.