Stoichiometry Part 1
Science
Let us learn more about Stoichiometry
All right, you chemistry studs and studded. We are ready to start chapter 12 second semester chemistry, chapter 12, look at the title here, stoichiometry. What the heck is stoichiometry? Well, stoichiometry is a process. Stoichiometry is a skill that you're going to need to use. This semester when you start taking a look at the next level of chemical reactions. Not only are we now going to take chemical reactions right the reactants write the products balance those reactions, but what we're going to be doing is we're now going to be taking a look at amounts. How much reactant and how much product are we going to have in a chemical reaction? So what are you going to need today? Well, these are notes. And so you're going to need obviously some paper to take some notes. You're going to need a pattern pencil to write this down, but folks, this is not a lecture where I'm going to be talking and you're just going to be listening. You're also going to need a calculator and a periodic table.
So if you need to right now, go ahead and pause me. And I'm sure you wanted to be able to do that for a long time now, but go ahead, pause me, get out your paper, your pencil, your calculator, and find a periodic table, and we will be ready to go. And here we go. So, you ready? Stoichiometry. Let's take a look, start with a definition. Stoichiometry is a process. It's a skill, a step by step, for determining how much the amount of reactants and the amount of products that we need for a chemical reaction to take place. Let's take a look at this chemical reaction that we have here. Nitrogen gas, reacting with hydrogen gas to form ammonia, ammonia vapor, gas. One of the things that we see right away is that this reaction is not balanced. So take a moment and balance this chemical reaction. So now we actually see that we've got one nitrogen combining with three hydrogens to produce two ammonias. What does this really mean? Well, one of the ways that we can interpret this is that one nitrogen. Molecule.
Can combine with three hydrogen molecules to produce two ammonia. Molecules. So one interpretation of this chemical reaction is that one molecule of nitrogen, combines with three molecules of hydrogen to produce two molecules of ammonia. Another way that we could see that is, instead of one, three and two, why couldn't it be two molecules of nitrogen? Combining with how many? Oh yeah. 6 hydrogen molecules to form four ammonia molecules. So we can see it that way. I can take it another step. Ten nitrogens. Plus 30 hydrogens. Make 20 ammonias in terms of molecules. What if I had 6.02 times ten to the 23rd, nitrogen molecules? How many hydrogen molecules would I need? While three times that many. And how many ammonia molecules would I produce? Well, two times 6.02 times ten to the 23rd. So here's the big a half folks. The big aha is that this is actually one mole of nitrogen. Combining with three moles. Of hydrogen to make two moles of ammonia.
Those coefficients, those things at the beginning, those coefficients in the balanced chemical reaction are not only particle coefficients, they're also and more importantly mole coefficients, so see that chemical reaction as one mole of nitrogen requires three moles of hydrogen to make two moles of ammonia. Big idea there folks really, really important. All right, so we're going to use this idea of mole ratios. To carry out chemical reactions to predict amounts of reactants and products, and there are 5 steps to make this happen. And those 5 steps are number one the first thing that you're always going to have to do is to write the balanced chemical equation. You have to know what the reactants are. You have to know what the products are, and you have to know what those coefficients, those numbers out front are, if you're going to write this reaction correctly. Step number two, and this is all going to make sense as we start going through it.
Number two, we're going to identify what's given in the problem the given information, gosh, that sounds a little bit like dimensional analysis, identify the given information, write it down and most importantly make sure that we include the units that we're talking about whether their atoms or moles or grams make sure make sure make sure that you label your given information. And then we're going to use dimensional analysis. And the idea here is we're going to always use dimensional analysis to convert that given information. From whatever units that we started with, to moles. We're always going to moles. Step number four, we're going to then use the mole ratio from the balanced reaction, like in the example that we looked at just a moment of go. Though that mole ratio one to three to two, we're going to use that mole ratio to convert from the substance that we started with are given information to moles of the substance that we're interested in. What we want to go to when we're all done, and then finally step number 5 and I'll move that up so you can see it is we're going to use dimensional analysis to convert from moles to the units that were interested in. So let's take a look and see what this really looks like.
Okay, so here's our example problem. If you have 10.5 grams of nitrogen gas and plenty and excess of hydrogen gas, we're not going to worry about the hydrogen gas right now. We know it has to be there in the problem, but we're going to ignore it for right now. But if we start with 10.5 grams of nitrogen gas. How many grams of ammonia are you going to make? Step number one to do this correctly the first thing that we have to have is we have to have a balanced chemical equation, so we have nitrogen gas and two reacting with hydrogen gas, H two to form NH3, and just a couple minutes ago we balanced this reaction and we found out that we needed one mole of nitrogen and three moles of hydrogen to make two moles of ammonia, so step number one is done now. We've written our balanced chemical reaction. Okay, so step number two, it says to identify the given information, we're going to write it down and we're going to make sure that we include the units. So let's go up to our problem, are given information is that we have 10.5 grams of nitrogen gas. So ten .5 grams of nitrogen gas.
You need to make sure that you do two things, that you put the label of the units that you're interested in, but also importantly what substance that you're in right now, because right now you're in the substance nitrogen, but when we're all done, we want to know how much ammonia we're going to switch substances here. And that's going to be a big idea. That's going to be a really big deal. So we just completed step number two, we identified the given information we wrote it down and we made sure that we included the units. So let's go to step number three. Step number three says to use dimensional analysis to convert that given information to moles. So we have our 10.5 grams of nitrogen, and we're going to convert grams of nitrogen to moles. Of nitrogen. How the heck do we do that? What's the piece of information that we know that allows us to convert from the units of grams to the units of moles? Well, that piece of information is the molar mass. So you need to go to your periodic table. Let's look up let's find nitrogen.
One mole of nitrogen. Well, the molar mass for nitrogen on the periodic table is 1401. But this is N two, so rather than 1401, we're going to write down 28 .02. So at this point, we've taken our grams of nitrogen, and we've converted it to moles of nitrogen, our grams, cancel out, were now in moles of nitrogen. So that's step three. Let's take a look at step number four. Step number four says to use the mole ratio from the balanced chemical reaction to convert moles of your given substance to moles of the substance that you're interested in. Okay, let's take a look at that again. We want to convert the moles that we're in right now, which is moles of nitrogen. And we want to convert that to moles of the substance that we're interested in. Let's go back up to the top and take a look at this. What do we want to go to? We want to go to a ammonia, so the substance that we're interested in when we're all done here is not nitrogen, but ammonia.
So using step four here, we're going to take our moles of nitrogen and convert it to moles of the substance we're interested in, which is the ammonia. Okay, so now I have moles of nitrogen to moles of ammonia. Where do I get the numbers that I need to put in here to make that mole ratio true to make that mole ratio correct? Well, it says to use the mole ratios from the balanced chemical reaction. Let's look at that balance chemical reaction. Our moles of nitrogen, one mole of nitrogen, to two moles of ammonia. So for every one mole of nitrogen, we need two moles of ammonia. And now our moles of nitrogen cancel out. The unit that we're in right now is moles of ammonia. So finally the last thing that we need to do is we need to use dimensional analysis to convert to the unit the units that were interested in. Not just the substance that we're interested in, but also the units that we're interested in, and again, we need to go back up to our original reaction, take a look what it says, and it says, we want when we're all done grams of ammonia. Okay? So what are we in currently? We're in moles of ammonia, so let's go ahead and using what we know.
We're going to cancel out moles of ammonia. And we're going to go to grams of ammonia. And again, how do we do that? We look on our periodic table. And we say, what's the molar mass of nitrogen? It's 1401. What's the molar mass of hydrogen? Well, it's one O one, but I've got three of those. So plus three O three, that's going to be 17 .04. All right, now we've canceled out moles of ammonia, and we are in the units of grams of ammonia, all you need to do at this point is you need to do the math. You're going to take your 10.5 grams, divided by 20.2802 times two technically divided by one, but we're not going to worry about that. Times 1704, and when we're all done, let me grab a calculator here quick. I'm going to take 10.5. Divided by 28.02. Times two, times 1704. And if I did my math right, rounding to three significant digits, I'm going to get 12.8 grams of ammonia. Produced. And there's the process. There's the process that you need to follow. All right. So now, you try, I want you to pause me at this point. And you try to do this problem.
Follow the 5 steps that I gave you. But go ahead and pause me. What I'm going to do is I'm just going to keep talking here, but I want you to pause me right now. The first thing I'm going to do is to write that balanced reaction. So if you get to the point where you're writing that balanced reaction and you get stuck on that, you can turn me back on, and you can get that balanced reaction, and then you can go from there. So go ahead and give me a pause. All right, so we're going to decompose nitrogen dioxide. So, and to O 5, decomposes into nitrogen dioxide, which is NO2. Plus oxygen gas, which is O2. Granted, I should put the states of matter in here. I'm going to leave them off for now just four convenient sake. So in balancing this reaction, I'm going to need a two in front of that. I'm going to need a four in front of that. That gives me ten oxygens. I've got 8 in the end O two, and two more makes ten. So that looks to me like the balanced chemical reaction.
So, here we go. The stoichiometry of this. I'm going to start with my given information by giving information is that I've got 5.8 grams of the N two O 5. And I need to convert, as it says in step number two, from grams of N two O 5, sorry about that. Two moles. Of N two O 5. And one mole of N two O 5, again, I got to grab my calculator, nitrogen way is 1401 times two plus 16 times 5, I get a 108.02. One zero 8 point zero two. All right, so grams of N two O 5 cancel. I'm in moles of N two O 5 at this point I want to convert from the given information the substance which is the N two O 5 to the substance that is asking about, in its asking about oxygen gas. So let's go. Moles? Of N two O 5. Two moles of O2. Looking at the balance to reaction, it's two moles of N two O 5. To one mole of O two? Moles of N two O 5 cancel out. I'm in moles of O two, but what they want are grams about two, so let's go from moles of O2. To grams of O2, oxygen weighs 16.00 times two is 32.00. Having a little problem with my pen here, so here we go. The math 5.8 divided by one O 8 zero two divided by two times 32, gives me and my answer for this to two significant digits because this has two significant digits would be zero point 8 6 grams of O2. Moles cancel out. 0.86 grams of O two, and there's the answer.
How'd you do? Did you get that one right? Did you get it right to the correct number of significant digits? I hope so. If not, if you're struggling, rewatch this again, take a look, walk through the steps a second time. It's perfectly okay to do that. I'm going to go ahead and move on to the next one. Okay, so this one's just a little bit different. Let's take a look at it again, and it says, when hydrogen and oxygen gas it actually probably should say when hydrogen and oxygen gases react together, they produce water vapor. If you started with 1.9 grams of hydrogen gas, how many water molecules would you produce? Oh, so now we're not talking about grams. Now we're actually looking at molecules. So, first thing I've got to do is I've got to write that balanced reaction. So hydrogen gas. And we know it's diatomic, plus oxygen gas, all so diatomic, are going to react together to produce water. Again, it's perfectly okay if you want to pause me at this point and try this on your own. Please do so if you are ready for this, go ahead and do it and then zip right on through it if you're not quite ready for it, then just walk right through it with me, so I'm going to balance that. It looks like I'm going to have a two or one and a two.
So there's my balanced reaction. So starting with 1.9 grams of hydrogen gas. So 1.9 grams of hydrogen gas. And I'm going to convert my grams of hydrogen to moles. One mole of hydrogen gas is 2.02 grams. Now that I have grams of hydrogen gas cancel out, I'm in moles. I want to do my mole ratio to go from the substance that I started with, which is the hydrogen gas, to the substance that I'm interested in, even though it says water molecules I still am going to use that mole ratio. I'm going to go from moles of hydrogen gas. To moles of water. Looking at my balanced reaction, that's at two to two. You can write it as two to two. You can write it as one to one, doesn't make any difference to me. It all works out in the end. So I'm just going to leave it as it is. Moles of hydrogen gas now cancel out. I'm in moles of water. But I don't want moles of water. I want molecules of water. So moles of water to molecules. What the heck is that relationship? Well, let's take a look at this. This should look familiar to you from the last chapter that we did. How do we go from one unit to another? Well, one mole, if I have one mole of anything, what do I know? Well, one mole of anything has how many particles. It has 6.02 times ten to the 23rd particles.
Now that could be atoms, if it's an element like iron, it could be molecules if we're talking about water. And sure enough, that's what we happen to be talking about so one mole of water molecules contains 6.02 times ten to the 23rd water molecules, so let's go back and put that in our problem. One mole of water has, and I'm going to have to squeeze this in here. 6.02 times ten to the 23rd. Molecules. And that's what I wanted. I wanted molecules. And that's what unit I'm in right now. So 1.9. Divided by 2.02, times two divided by two or times one divided by one, times 6.02 times ten to the 23rd, and we wind up with to the correct number of significant digits, which is two, 5.7, times ten to the 23rd, water molecules. Bam. There you go. How'd you do? Are you getting this? Once you get it, it's pretty straightforward because it's the same process every single time. If you're not getting it, like I said, go back and watch it again. Go back and watch it, rewatch it. Make sure you write down your questions so that when you come to class, too, you can stop me and say, hey, you know, everything made sense up to this point, and I'm totally lost here, misses Wolfe, what do I do now? All right, so we took a look at that. Let's go ahead and move on to another one where you get to try it again. And this one, again, read through it and pause me, see how you do. When aluminum metal is placed into a solution of hydrochloric acid a reaction occurs, write the reaction predict the products of this reaction, okay? So that's the first thing you need to do. Go ahead, pause me.
I'm going to keep on talking, and when you are done with this problem, then pick up where you left off here and see how you did. All right, so here's my reaction. I'm going to react aluminum metal with hydrochloric acid. Pulling information that you learned in chapter ten. What type of reaction does this look like? The aluminum is an element, and here we have a compound, the aluminum is going to kick out the hydrogen. We're going to get aluminum chloride. Now when I put this together, I need to think about my charges. Look on your periodic table, aluminum is plus three, chlorines minus one, so aluminum chloride is going to be ALCL three, plus the hydrogen is going to be left by itself, but we of course know that hydrogen can not exist just as H, it's one of the 7 diatomics, so there's H two. So there's our reaction, but it's not balanced. So we need to go back and we need to balance this thing. All right. If I put a three in front of the HCL, I'm going to wind up with a problem because I'm going to have an odd number of hydrogens, which is not going to work.
So I'm going to start by putting a two there in front of the aluminum chloride that gives me 6 chlorines. I can put a 6 here in front of the HCL. Therefore, I'm going to need three hydrogen gases and two aluminums. There's my balanced reaction. So I have my reaction balance. Now it's time to do the stoichiometry. All right, so we're starting with 2.9 times ten to the 23rd atoms of aluminum. 2.9 times ten to the 23rd, atoms of aluminum. Okay, this is a little bit different again, because we're starting with atoms instead of grams, but that's okay. What do we always have to go to? We always have to go to moles because that's the only way we're going to use our mole ratio. So atoms to moles one more has 6.02 times ten to the 23rd. Atoms. Atoms of aluminum cancel out were in moles. Now we can use our mole ratio moles of aluminum to moles of what do we want to know? We want to know hydrogen gas. So we want to go to moles of hydrogen. And looking at our balanced chemical reaction, two moles of hydrogen, will give us three moles of hydrogen.
And so we're going to cancel out our moles of aluminum, and finally we don't want moles of hydrogen, but instead we want grams of hydrogen, so moles of hydrogen to grams of hydrogen, using the molar mass, one mole of hydrogen gas weighs 2.02, and when I do my math here, 2.9 times ten to the 23rd, divided by 6.02 times ten to the 23rd, times three divided by two, times 2.02, I get my answer of, and how many sig figs two sig figs, 1.5, times whoops, not times anything. 1.5 grams. Of hydrogen gas. There you have it, folks. There's the introduction to stoichiometry. At this point you need to go back to your checklist, take a look at what your next step is, and if you got stuck on some stuff along the way, make sure you ask some questions otherwise good. Luck. Go writers.