Hello, this is a lecture topic video discussing chemical equations. So at this point in the semester, we have a great description of the atom, and we've talked about how atoms can come together to give us some compounds. Now what we want to do is start to describe how different compounds or elements can come together and react with one another. And the way we kind of communicate that to an audience is through chemical equations.
But let's remember all the way back to the first module where we talked about... John Dalton. He's considered the father of modern day chemistry.
He came up with an atomic theory. And he had three postulates. One of them was matter was composed of these tiny indivisible particles called atoms.
Now we know that they're not indivisible because atoms have electrons. And within that nucleus, we have protons and neutrons. But generally speaking, we have these atoms and all atoms of a given element are identical. That was John Dalton's essentially his first postulate of his atomic theory.
The second one was compounds are made up of specific combinations of atoms of two or more different elements. And then that led us to the third postulate, which is what we want to talk about today. A chemical change, all right, or we can call this a chemical reaction, involves a reorganization of the atoms in one or more substances, all right, and we communicate these changes through chemical equations, all right? These chemical changes we can communicate by using chemical equations, which are essentially a representation of a chemical reaction, all right? And what they do is they show us the relative numbers of our reactants in our product molecules, all right?
So what do I mean? So down here, what you can see is we have a chemical equation where we're reacting carbon dioxide, all right, with water to give us this compound C6H12. H12O6, all right, and oxygen, okay?
And so what we can do is we can start to read this like a sentence. And so where we're talking about our reactants, all right, that's our CO2 and our H2O reacting with one another, all right, to yield, okay, which is kind of shown by this little arrow symbol here, our products, which again here is C6H12O6 and diatomic oxygen, okay? And so what we can... kind of do is break this up into three parts, all right?
The left side, okay, so again, this is all relative to our arrow, all right, which is again showing us essentially the statement we are yielding something, all right? On the left side, we have our reactants, okay? These are the starting substances in any chemical equation, all right? And again, they're always on the left-hand side of the arrow in a chemical equation, all right?
And then when those two reactants react together, They yield something. And that's our products. And those products, again, being the product of the reaction of our reactants with one another, those are always shown to the right-hand side of an arrow in a chemical equation.
And so there's a lot of important things that we can see from our chemical equation, our chemical equation. We can see exactly what our reactants are. All right. And you can see here again, CO2.
And H2O, those are our reactants, all right? And we represent those using our chemical formulas, all right? So our chemical formulas. And remember, our chemical formulas are kind of showing the relative ratios, all right, in this case of carbon and oxygen coming together to give us a given compound, all right? So our chemical formulas, all right, are showing us our relative ratios of the atoms of elements within a given compound, okay?
And so again, you can see we have our carbon dioxide, which reacts with water, all right, again, to give us our C6H12O6 and our diatomic oxygen. Those are all of our chemical formulas of the compounds that are involved with this reaction, okay? Then what we can see is our physical states of matter, okay?
So is it a gas, all right? So CO2 is a gas. Or is it a liquid, all right, in the case of water within this given reaction, all right? Or is it a solid? Okay.
So our C6H12O6. All right. Again, which this is just a carbohydrate. I guess I should have said that before. It's just a carbohydrate.
So it's a sugar. All right. And in terms of our diatomic oxygen, all right, it is also a gas.
So. What is the physical state of our different compounds? All right. And as you'll see, this is actually really quite important. The other really important thing that we see here are these, what we call our stoichiometric coefficients.
All right, so stoke geometric coefficients, all right, that are these numbers, okay, these numbers outside, okay, on the left-hand side of a given compound, all right. Again, this is telling us how many molecules of each reactant, okay, must come together to yield the listed number of the product molecules. So again, it's telling us the number of molecules that we have within our given chemical equation. It's telling us the ratio of how these compounds need to react to give us the products of our reaction.
In other words, to make this reaction work, we need at least six molecules of carbon dioxide to react with six molecules of water, which gives us one. All right. One molecule of sugar plus our six molecules of diatomic oxygen. OK, now you see here that I wrote in this one.
All right. Generally, we do not do that. OK, if there is no coefficient, all right, stoichiometric coefficient out in front of a given compound, that means there is only one of those compounds.
And by convention, we normally do not write the number one. OK, so chemical reactions. All right. So what you'll see is that we have five general types that we'll be talking about, okay, this semester. And those types are listed here.
So we have combination or synthesis reactions. That's where we have two or more of our reactants, so in this case A and B, all right, that come together to give us a single product. That, again, is called a combination or synthesis reaction.
So two or more reactants. giving us a single product. Then we could also have decomposition reactions. Well, in this case, this is where we have a single reactant.
And what it does is it decomposes, it breaks down into two or more products. Okay. So in this case, A and B breaks down to give us free A and free B. And then we have single replacement reactions. In this case, this is where, if we're looking at this general scheme here, this general chemical reaction, all right, what you can see as we shift from our products to our reactants, A has essentially replaced B in terms of its ability to react with comp or comp.
molecule C or element C, okay? So in this case, this is a single replacement reaction. A plus BC gives us AC plus B, all right?
One of those, one of our reactants has replaced another within a given compound, okay? Versus a double replacement reaction, all right? That's our fourth type of chemical reaction that we'll see, all right? Is where essentially our partners within these compounds swap positions.
All right. So in other words, as we look from our reactants to our products over on this side, we have an AB compound plus a CD compound. And when we look at our products, what we see is that A and C have swapped with each other. OK, so this is called a double replacement reaction. OK.
And then finally, what we'll see are. Combustion reactions. All right. Where we have our fuel. All right.
You always have to have a fuel. And we're talking about combustion reactions. And then we need a source of oxygen. In this case, it's diatomic oxygen gas.
And then we need us essentially a source. some way, some source of energy to get these guys to react with each other. And what do we always get when we're talking about combustion reactions? We always get carbon dioxide, and this is a gas, all right?
And we usually get water or always get water, okay? Those are always the products of a combustion reaction, all right? You have to have a fuel, all right?
And it's reacting with diatomic oxygen, and it always gives us CO2 gas plus H2O liquid, all right? But... The fact of the matter is these reactions generate a lot of heat.
Okay, that's what this is, right? So generally our water is vaporized into its gaseous form. So generally you get H2O in its gaseous form. So this is our fifth type of reaction, a combustion reaction. All right.
So now, again, when we're looking at these chemical equations, all right, it gives us two really important types of information. One, again, is the nature of the reactants and products. Again, what do I mean by the nature of a reactant or a product?
We're talking about its physical state. All right. Is it a solid?
Is it a liquid? Or is it a gas? All right.
And again, the symbols for those. is just solid. And notice that they're always in parentheses or lowercase L for liquid or G, all right, lowercase G for our gas. Now, another state of matter, okay, or physical state, because we see this so often, like chemistry happening in an aqueous solution. And what do I mean by that?
In water, all right, so reactions happening in water. Remember, if you're thinking about our bodies, all right, all of our reactions are happening in a water environment, all right? Because again, we are 70, 80% water, all right, as being human beings, okay? So all the reactions inside of our body are happening in an environment of water.
This is called an aqueous solution, okay? So whenever you see something that is essentially dissolved in water and then carrying out some reaction with something else, it is in an aqueous state, okay? So for example... All right.
We can look at this chemical equation below this table where we have hydrochloric acid. Again, what's our symbol for an aqueous solution is that lowercase aq in parentheses. So that's telling us that hydrochloric acid or that hydrogen chloride has been dissolved in water and it's reacting with solid sodium bicarbonate.
All right. That's what this is telling us. And what do we get when we do this reaction? We get.
gaseous carbon dioxide plus liquid, all right, water, H2O, and a salt, sodium chloride. Again, and this is going to remain in an aqueous state, all right? So that's why it has the AQ.
Now, our states of matter are really, really, really important, all right? Because sometimes a given compound has to be in a particular state of matter in order to react. For example, gasoline, all right?
We all know that gasoline is a flammable substance, but in fact, liquid gasoline is not, all right? So in order for gasoline to actually be flammable or be able to be combusted, all right, it has to actually be in its gaseous form, okay? So again, our states of matter, when we're talking about a given reaction, are really, really, really quite important, all right?
The other thing that these balanced or chemical equations tell us, right, is the relative numbers of each compound that's, again, reacting, all right, and what's being produced, all right? And so, in order for this to work, all right, we have to be working with, again, I kind of cut to the chase a second ago, balanced chemical equations. We have to have a balanced chemical equation, all right?
In other words, what do I mean by that, all right? The number of atoms of a given element OK, on the product side, on the left hand side of our reaction arrow have to equal the number of elements or atoms of elements on our product side. OK, they have to be balanced.
And why is that? Well, the first law of thermodynamics tells us that we can neither create nor destroy matter. All right. But on the other hand, we can change its form. OK, so in this case, on our product side.
we have carbon and hydrogen and oxygen gas, all right, forming these molecules, methane, all right, and we have diatomic oxygen, but on the other side, all right, of our balanced chemical equation, all right, we now have carbon, oxygen, hydrogen, but they're in two different forms, all right, so now they're in carbon dioxide and water, but if you count up the atoms of a given element within the products relative to the same element on the left-hand side and our product side, they are equal to each other. So let's dive into this just a little bit more. So again, we have methane reacting with diatomic oxygen, okay, and giving us our CO2 gas plus our H2O gas, all right? Again, what kind of reaction is this? This is a combustion reaction.
All right. And so what you can see is from this balanced, okay, this balanced. chemical equation, all right, it tells us that we have one molecule, again, the one's not indicated with our coefficient, our stoichiometric coefficient, all right, that reacts with two molecules, right, so here's one, here's two, of diatomic oxygen gas, and what results from this, again, we can read this like a sentence, okay, methane reacts with oxygen gas, two yield, okay, two yield, carbon dioxide gas, one molecule of carbon dioxide gas. Again, remember, stoichiometric coefficient of one is not shown, all right, plus two molecules of H2O.
So again, here's one and here's two. Now, how do I know that this is balanced? Well, again, what can we do?
We can start to tally up the atoms of the different elements on either side of our reaction arrow. So if we're looking at our products or our reactants, excuse me, we have one carbon atom coming from methane and we have four hydrogen atoms coming from methane. All right.
And then if we look at our water or excuse me, our diatomic oxygen gas, what do we get from there? Well, again, diatomic oxygen gas has two oxygen atoms bound together. All right. And this stoichiometric coefficient is telling us that there are two of them. So therefore, we have a total of four.
oxygen atoms. Okay. Now, if we look at our product side, all right, again, what do we have? We have from our carbon dioxide, we have a single carbon atom.
All right. And we have two atoms of oxygen. Okay.
So one and two now from our water, all right, water's H2O that tells us for every oxygen, we have two hydrogens. All right. So, and we have a stoichiometric coefficient of two.
Therefore, we have our four hydrogens, okay? And we have two additional oxygens. So there's oxygen three and oxygen four. If you look at the number of atoms of a given element on our reactant side versus our product side, you now can see very clearly that they are equal to each other. This is a balanced chemical equation.
This does not violate The first law of thermodynamics that says we can neither create nor destroy matter, okay? Matter is conserved, all right? But it can change forms. Again, we have different compounds on our reactant sides relative to the product side.
Okay, so the first thing we want to do when we're thinking about balancing chemical equations and writing them, all right, is we have to determine what reaction. is occurring. All right.
Again, chemistry is cumulative in the fact that we expect you to be able to do things that we've learned in the past and be able to apply them in the present. Okay. So a lot of times these reactions will be written out in words.
So you need to be able to take chemical names or compound names and be able to convert them into their chemical formulas. All right. But what you need to do is we need to figure out what are the reactants and what are the products. All right.
And then again, we have to think about what the physical states are, like what are the physical states of our reactants in our products, because that is one of the most important things about a balanced chemical equation. OK, so that's our first step. What reaction is actually occurring and what are the states of our reactants and products? Second step, right, is you want to write an unbalanced equation that summarizes what we determined in step one.
OK. And then after that, what we need to do is to start to balance it. We need to determine those coefficients, the stoichiometric coefficients that are necessary. So the same number of each type of atom appears both on the reactants and the product sites. Now, an important thing here is that you do not want to change the identities.
All right. In other words, you cannot change the formulas of any of our reactants and products in order to lead to a... balanced chemical equation because as soon as you change the formula of either a reactant or a product, you have a completely different chemical equation, all right? A different type of reaction that may be occurring.
So we never do that. We always balance, all right, by essentially adjusting our stoichiometric coefficients, okay? So generally, what are the steps, all right? There really are no real basic steps that you should follow every certain time when you're doing this.
All right. But there's some general things that you can do to help yourself out. Generally, you want to balance compounds first.
So like carbon dioxide, for example, is a compound. So before balancing elements or compounds that are consisting of a single element like oxygen gas. Okay.
So diatomic oxygen. All right. And generally you want to look at the most complicated molecules first.
Is there a way? to balance those atoms, all right, that are within those most complicated molecules first. Now, and something that will really, really help you is this second kind of general rule here, all right? Remember your polyatomic ions, all right, whether it be a polyatomic anion or a polyatomic cation. You do not want to change those from one side to the other.
In other words, if you have chlorate, okay, on your reactant side, all right, and chlorate ends up in a product, all right, don't change the formula of chlorate. Don't break it up. All right.
Keep it as a group because the benefits of doing that is that then, all right, you can balance it as a group. Okay. So instead of breaking chlorate into a single chlorine and three oxygen atoms, all right, we can just balance chlorate as a group. All right. So one chlorate, all right, on the reactant side should ultimately end up.
with a single chlorate on the product side if you have a balanced chemical reaction. Or maybe it's two versus two on the product side or three, so on and so forth. So balance them as a group, all right?
It'll make your life that much easier. So here's a sample problem. Solid iron three sulfide reacts.
Again, pay attention to the wording, right? Because we can read a chemical equation as a sentence essentially. All right, so we have solid iron three sulfide reacts. interacting with gaseous hydrogen chloride to form, remember to form, that's telling us that that's our reaction arrow, solid iron three chloride and hydrogen sulfide gas.
So the question is asking us to write a balanced chemical equation. And so again, we are expecting you to be able to take the names of a compound and be able to write the formula for the compound. So iron three sulfide, all right.
So that's Fe2S3. And they're telling us it's a solid. So let's not forget our state of matter, all right, reacts with gaseous hydrogen chloride.
So that's HCl. And again, it's a gas. And notice I'm leaving spaces out in front of these guys so that we can add in some stoichiometric coefficients. Again, because the first step is let's just figure out and write out the... chemical equation that's actually happening.
So generally what's going on. All right. And then it reacts to form.
These two guys react to form. That's our arrow, our reaction arrow, solid iron three chloride. So F E C L three.
And again, it's a solid and hydrogen sulfide gas. So H two S and it's a gas. All right.
So again, go. back, review the rules for naming, all right, and how the names of a given compound can help you determine what the formula is for that given compound. Because again, it's really, really important.
Sometimes these formulas for these different compounds will be given, but other times they may not be. All right. So now what do we need to do?
We know the general equation, all right? We have a very general chemical equation, but it's clearly not balanced. So the first thing we have to do is tally the different numbers of atoms of a given element on each side of the equation. And so normally what I do is I just write a little table where I put the different elements in the middle. So we have iron we're dealing with, sulfur we're dealing with, hydrogen, and chloride.
And then I do in my initial tally, how many of these elements are on the reactant side versus the product side. And what you can see is that we have two irons, okay, on a reactant side, we have three. sulfurs, we have one hydrogen, and we have one chlorine. Now what's going on on the other side? Well, in this case, we have one iron, one sulfur, two hydrogens, and three chlorides.
Okay, so now what do we want to do? We want to tackle, generally speaking, our compounds first. Well, all of these guys are, are compounds made up of different numbers of different elements. All right. So we don't have any pure element, all right, or any diatomic molecule or compound that is made up of a single.
element. So like diatomic oxygen, okay. For example, or maybe even ozone that's O3.
All right. And so, but we can, what we can do is we can look and see what the most complicated thing is. Okay. And so for me, I would say that that's our iron three sulfide.
It looks to be the most complicated. So is there a very straightforward way where we can start to balance out? All right. Those elements within iron three sulfide.
All right. And so we can look at our tally, the difference between our irons. You focus on one type of element first.
All right. So let's focus on the irons. We have two on our reactant side. Let's just write that here. Reactant products.
All right. So we have two on our reactant side versus one on our product side. So the very easy thing for us to do is to add a stoichiometric coefficient of two in front of our iron three. chloride. All right.
Well now how does that change our tally? So again, as you start to add these coefficients, keep track of the number of atoms of a given element. So keep track of your elemental tally is what I call it. All right.
Well now what do we have? We have, all right, two irons and again iron three chloride FECL3 that tells us for every iron there are three chlorides or three chlorine atoms. And so now that we have that two out front tells us there are two of these molecules. All right.
And so that tells us that now we have six chlorines. So now we have six chlorines. All right.
So we have successfully balanced out our irons in this case. Now, what's the next thing that's a part of our iron three sulfide? Again, our original molecule.
All right. It's the sulfur. So why don't we try and balance out our sulfurs?
All right. Well, we have three on our reactant side and a single one on our product side. So let's add a three out in front of our hydrogen sulfide gas. All right.
Okay. Well, what happens when we do that? Because again, we want to keep tally, right?
Well, now hydrogen sulfide gas is H2S, two atoms of hydrogen for every atom of sulfur. And now we have three of these molecules. So essentially what we've gone from in this case.
is now instead of having two hydrogens, all right, we now have six hydrogens, all right? And now instead of a single sulfur, we have three sulfurs. So if we're keeping track of our tally, what we can see is that our irons are done.
So we can just circle that, all right? We got two on our reactants, two on our product side. Our sulfurs are now done. We have three on our reactants and...
three on our product side. And now what we see is that we have to deal with our hydrogens and our chlorines. Okay. And so we don't want to come back over here. All right.
Again, what you'll see is sometimes you'll have an ability to maybe want to place something in front of iron two or iron three sulfide. All right. We don't want to do that in this case.
All right. Because our irons and our sulfurs are completely balanced. All right. And it wouldn't help us in the first place.
Okay. So, but what we want to do, all right, is we want to focus on that, uh, that hydrogen chloride, that, um, hydrochloric acid. Okay.
And again, what's nice. What do we see on our product side? We have six hydrogens and six, uh, chlorines.
All right. That need to be balanced on our reactant side. Well, what's nice is that on our reactant side, okay. Our hydrochloric acid, all right.
Or our hydrogen chloride come together in a one-to-one ratio. There's one atom of hydrogen for every atom of chlorine. So the very easy thing for us to do in this case, all right, is to add a six out in front.
And what does that do? That takes us now to having six hydrogens and six chlorines. And so now we have essentially finished balancing this chemical equation, regardless of whichever side of our chemical equation, which again, we're talking about.
either side of our reaction arrow, we look at the given elements in this chemical equation are balanced. All right. We have two atoms of iron on either side, three atoms of sulfur on either side, six atoms of hydrogen on either side, and six atoms of chlorine on either side. Matter is neither created nor destroyed.
All right. So we're not violating the first law of thermodynamics. And you can see, because it's allowed. that our matter has just merely changed its form. We've made new compounds from the things that we started with.
All right. So that's that question. Let's take an example of the combustion of liquid ethanol.
So the combustion of liquid ethanol, remember, that's one of our five types of reactions. And again, you'll see when we start talking about some compound names where we use maybe common names instead of the defined names that you can get from like. going through our rules of naming, we're going to give you the formulas for those compounds. We wouldn't expect you to know, all right, the formula for ethanol. So again, that's something we would give you.
All right. So take some reassurance in that. Okay. But we have a combustion reaction.
And remember, a combustion reaction needs oxygen. All right. So we need oxygen. Okay. And what does it do?
It forms or it yields, right? This is telling us this is our reaction area, carbon dioxide. and water vapor.
And so what we want to do is we want to write a balanced chemical equation. Well, we can already see, all right, that we're dealing with ethanol. That formula was given to us.
So we can take CH3CH2OH, all right, and we can react it with oxygen. Now you may be tempted to write oxygen, all right, and again, this was liquid ethanol. So let's not forget our phases.
Okay. Those are really important. And we know that oxygen is a gas.
Well, remember there are some elements that love to form diatomic versions of themselves and oxygen is one of those. All right. And in fact, all right, there are seven of these things.
So if we look, how do you know which ones are, they is actually fairly straightforward to know. So let's get a highlighter here. All right. Our nitrogen. Okay.
Loves to form. a diatomic molecule with itself? Our oxygen, our fluorine, our chlorine, our bromine, and our iodine. All right.
So right there, that is six of them. And what you'll say, hey, Dr. W, you said there were seven. All right. Well, the seventh one is hydrogen. Okay.
So these guys love to form diatomic molecules with themselves. So when you see a question saying gaseous oxygen, it's not. just a single O. All right.
It's not just oxygen in that sense. All right. It is diatomic oxygen.
Oops. There we go. Okay.
All right. And it forms again, this is a combustion reaction. Always get carbon dioxide gas.
and water vapor. And it's usually vaporized. It's usually in its gaseous form because of all the heat that's given off in one of these reactions.
Okay. So now what do we need to do? We need to balance, okay, our chemical equation.
All right. We need to come up with those stoichiometric coefficients that make it so that all of our carbons, our hydrogens, and our oxygens on the reactant side. Okay, equals those on the product side.
So we can do our initial tally, all right? In this case, we have two carbons on our reactant side, all right, six hydrogens and three oxygens. Whereas if we look at our product side, we have a single carbon, two hydrogens and three oxygens.
Okay. And so again, remember, we want to work with compounds. All right.
First that are made up of multiple different elements. Okay. Versus those compounds that are made up of a single element. So, and if I were you, okay, again, cause there's no right or wrong way to do this.
There's just general rules of thumb to help you along the way is I would leave this guy for last. Okay. Cause that's going to ultimately help us out. All right.
And what I would do is I would try to focus on working out a way so that we can start to balance those elements within ethanol. All right. It is probably, in my opinion, our most complex compound here in this chemical equation. So what we can see is first that we have two carbons on our reactant side and only a single one on our product side. So how are we going to start this?
All right. We can add a... documetric coefficient of two in front of our carbon dioxide gas. Now, how's that going to change our tally?
Because again, you want to keep track of your tally as you're starting to add these coefficients in. Well, that's going to take us now to having two different carbons. All right. And again, carbon dioxide, one carbon for two oxygen atoms. All right.
And now we have two molecules of this. So that tells us that now we have four oxygens coming from. the CO2, but we still only have one from the water.
So we now have five oxygens. Okay. All right.
So our carbons are now balanced. All right. Well, how about our hydrogens?
Well, we have six, okay. On our reactant side versus two on our product side. All right. So how is, is there a way for us to essentially balance that out?
Well, Again, the nice thing here is that the hydrogens are only involved with a single, all right, compound on our product side. All right. And again, water. There's two of them for every oxygen.
So what we can do is we can add a new coefficient out in front. And let's add a three. Okay.
And when we do that, all right, how does that change our tally? Well, now we now have six hydrogens. on our product side.
So now you can see our hydrogens. All right. Now have six, both on our reactant side and our product side. So they're essentially done.
All right. Well, how have we changed the tally for the oxygens? All right. Well, now we have three oxygens coming from our water and we had four coming from our carbon dioxide. So now we have a total of seven.
Okay. All right. So now it just looks like we really need to focus on those oxygens. Okay.
Well, we have seven on our... product side and only three on our reactant side. So we need to do something to the reactant side.
Now you may be again tempted to go start adding coefficients in front of ethanol, all right, because it contains an oxygen, all right. But this is where I'm going to tell you, don't do that. We've already balanced out most of those elements that are within, okay, that are within ethanol, all right.
So it's not the logical choice. If we add a stoichiometric coefficient out in front of ethanol, we're going to have to start all the way over at the beginning. Okay. So we're going to have to go back to the drawing board and start rebalancing everything.
All right. Now the nice thing here is that we have a compound that contains multiple atoms of a single element. All right.
And so how can we change that coefficient so that we essentially get seven oxygens on our reactant side, which would equal those seven oxygens on our product side. All right. Well, we have a single oxygen coming from ethanol. All right. So that tells us.
that from these O2 molecules, this molecule of gaseous oxygen, we're going to have to get six oxygens. And so the easiest thing for us to do here, of course, is to add a three out in front of our oxygen. And when we do that, now we have six oxygens coming from our gaseous oxygen, plus the one from ethanol.
And you can see now that we have seven. or seven atoms of oxygen on our reactant side, which matches those seven oxygens on our product side. So our oxygens are also balanced. So now what you can see is we have a balanced chemical equation.
One molecule of ethanol will combine with three molecules of diatomic oxygen to give us two molecules of carbon dioxide and three molecules of gaseous water. All right. This is telling us All right.
Again, the ratios of our reactants to the products. All right. It's telling us how many molecules of each of our reactants we need in order to get so many molecules of the compounds of our products. OK. And again, we haven't broken the first law of thermodynamics.
We have neither created nor destroyed matter, but our matter has changed forms. All right. Which is perfectly allowable. All right.
Now, some participation questions. The first one. is I've given you five reactions here, and what I want you to do, and they're all balanced, okay, each one of these are balanced, and I challenge you, take a closer look at these and see and make sure that this is in fact true, all right, but they should be balanced, all right, but what I want you to do is essentially tell me the type of chemical reactions each of these reactions represents, okay, so go back to that list of five, all right, that we talked about back on the second or third slide of today's lecture topic video.
Review what the different reactions are, all right, and then categorize these guys. The second one is write a balanced. chemical equation for the decomposition.
So here I've told you the type of reaction of liquid nitroglycerin to produce nitrogen gas, carbon dioxide, water vapor, and oxygen gas. All right. Remember those elements that love to make diatomic versions of themselves. All right.
That would be our nitrogen and our oxygen. So again, you have to start to keep these things in mind as you write. our chemical equations. Okay.
It's really important. If you just write plus nitrogen, all right, or plus oxygen, it's not going to work out for you. Okay. So again, remember that our nitrogen, oxygen, fluorine, chlorine, bromine, iodine, and hydrogen love to form diatomic versions of themselves.
Okay. But that's not what the question is asking for. So yes, you need to write this decomposition reaction.
You need to balance it, all right? But what the question is asking for is what is the coefficient for water, all right? When the equation is balanced using the lowest whole number coefficients, because again, that's what we're shooting for when we write our balanced chemical equations, all right? The stoichiometric coefficients should be, ultimately end up being whole numbers, okay?
So again, answer what the question is asking for. What is the coefficient for water? All right. When you have this balanced chemical equation.
And then the third question here in blue is write the imbalance, a chemical equation for the reaction of gaseous ammonia. All right. With oxygen gas to produce.
All right. Again, to produce that's telling us that's where our reaction arrow goes. Gaseous nitrogen monoxide and liquid water. Well, in this case, if you get your general reaction and you balance it.
All right. The question's asking for what is the sum, the sum of all the coefficients, okay, when the following equation is balanced? Again, using the lowest whole number coefficients. All right, the thing here is don't forget, if you have a single molecule of something, we don't indicate one in our balanced chemical equation, but it is assumed that the stoichiometric coefficient is one. All right, so don't forget to include that one within your tally.
All right, if you guys have questions. Don't hesitate to reach out. I'm happy to help.
And I hope you have a great rest of your afternoon and I'll be seeing you all soon. Take care.