SUPERENEM: Aula #81 | Química

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[00:00:07]Hey guys. Let's have another live class here with you all, and today we're going to talk about a topic that is, I would say, quite difficult for many people, but I hope we can unravel and develop it well, which is stoichiometric calculation, okay? So, I'm going to show you a series of examples of how we can try to make this subject simpler and more accessible. There will be some calculations to do, that's a fact, but look, you can be sure of one thing: stoichiometric calculations are a topic present in any college entrance exam you're going to take, okay? And obviously, the ENEM exam will be no different, right?

[00:00:51]So, I'm going to show you some cases of all kinds, okay? So that you can understand each case individually and that can help you solve problems, okay?

[00:01:04]I hope you can use today's examples as a model, okay? so you can deal with whatever issues come up. I'm looking at the chat on the other screen, I'm following along. If you have any questions, let me know. Don't forget that the second list of questions is available until Saturday, May 9th, okay? You can access it through superenem.com.br, and the list is available to those who have registered on the platform. Come on, let's look at this list, let's do the training, because good learning consists of this, right? Classes, training, all the time like that, so you can stay sharp, so you can learn a little bit about everything, so you can develop well. Very good. I'm just checking out who's watching me right now, and people are arriving, so we'll get started from here. Osaldo, who is from Calcaia. Very good.

[00:02:07]José Manuel is here too. Good afternoon to you all. Isa Suzi is here. Juliana Paulo, good afternoon. Good afternoon everybody. Okay, let's go, let's follow along so we can see what it's like, what we're going to learn in today's lesson.

[00:02:26]Stoichiometric calculations consist of determining the quantity of substances that participate in a process, a reaction, a sequence of reactions, and so on, okay? So it's important that before we proceed with the stoichiometric calculation, we understand a few things, a few ideas that we'll really need to use, which I call chemical quantities, such as atomic mass, molecular mass, mole, molar mass, and things like that. So, look, the first thing I want to show you is the atomic mass unit, represented by the letter U. A while back, this atomic mass unit was represented by UMA.

[00:03:11]Then it fell out of use and came to be represented by the letter U. Simpler, right? And it's the standard we use for measuring the masses of atoms and molecules.

[00:03:22]And it corresponds to one- twelfth of the mass of the carbon-12 atom.

[00:03:29]Note that this 12 is the mass number of the carbon atom, which is the most abundant carbon in nature. The most abundant type is carbon with a mass number of 12. Carbon-13 also exists, but in a rather small percentage.

[00:03:46]Despite this, it has its uses.

[00:03:49]Carbon-1 is used in nuclear magnetic resonance, which is that technique we use for medical diagnosis, and I don't know if you know, but magnetic resonance is used to discover the structure of a molecule, to know what the groups that compose, for example, an organic compound are, with those sometimes very complicated structures and so on.

[00:04:16]And furthermore, there's carbon-14, which is a radioactive carbon atom.

[00:04:21]It is a beta particle emitter and is used in determining the age of fossils. But this carbon-12 that I'm talking about here is stable carbon and it's very abundant.

[00:04:34]Given that carbon is a component of all living matter, because everything that is organic contains carbon, right?

[00:04:42]Therefore, given the presence of carbon in a myriad of different molecules, carbon is chosen as the standard of mass. This fraction of a 1/10th of carbon, therefore, is used as a standard for the atomic mass unit.

[00:04:58]Well, the atomic mass of an atom. It may seem redundant to say " atomic mass of an atom," but if you look at this screen, there's the atomic mass of the atom and there's the atomic mass of the element. Therefore, it's not the same thing. When I say the atomic mass of an atom, I'm thinking of the mass of an isotope, okay? a particular isotope, an isotope of that element, and we have a high- precision device to measure that mass, which we call a mass spectrograph, okay? This mass spectrograph works with samples of atoms that are deflected in magnetic fields. He is able to estimate the mass of atoms with good accuracy. I showed you here the example of the silicon atom, which has a mass number of 28, and look at its atomic mass, which is practically equal to 28, right? 28 is the mass number. When you look at the atomic mass, there it is, 27.98 units. I can guess, roughly, that atomic mass is almost 28, right? So we notice that there is a very close proximity, as well as looking at this iron atom here with a mass number of 56 and its atomic mass, notice, almost 56, right? Approximately 56, but not exactly. So, I mean, the values ​​are approximate, but this shows that there is a difference between the mass number, which is a whole number, 28, 56, but the atomic mass, which is a decimal number. If you round it off, you'll get a good approximation of the mass number. This one has approximately 28 units, this other one approximately 56, but they're not exactly the same, right?

[00:06:59]And the atomic mass of the element is the weighted average of the masses of its isotopes, using the relative abundances as weights for the average. In a weighted average, we need to assign percentages to the values we are going to average, which makes a weighted average different from an arithmetic average. In that case, I need to consider how many of each isotope I find in nature in order to use that information to determine the average atomic mass for that element. Then I'll give you the example of the element boron. As you can see, it occurs in nature as boron-10, which has an atomic mass of almost 10.01 units. It occurs in 20%.

[00:07:49]While the other boron atom that exists in nature, boron-11, has an atomic mass of 11.01, which is also almost 11, right? And 80% corresponds to him.

[00:08:01]This means that of all boron atoms existing in nature, 20% of these atoms have a mass number of 10. The remaining 80% have a mass number of 11. And I want to know what the atomic mass of boron is. So, what calculation do I need to make? That's because this atomic mass will be 20% times the mass of the first 10.01, but I 'll round it off, there's no problem rounding to 10.80%, which is the percentage of the other one, right? Times 11 units. I'm rounding it off now too. 11.01, I rounded it up to 11, right? So, take a look at how this turns out. If I take 20% of 10, what will that be? It'll be two, right? That'll be two, two units.

[00:08:58]So, I'm left with two units plus 80% of 11, which gives me 8.8, right? So this leads me to the atomic mass of the element boron, which would be 10.8 units.

[00:09:16]This value of 10.8 units for the atomic mass of the element boron is what you find in the periodic table. When you take a look at the element, you'll find the symbol for boron there, okay? Over there in family 15, in family 5A, actually, sorry, sorry, in family 13, family 3A, family 15 is nitrogen. So you can find boron there in group 3A, in the current numbering, group 13, right? Group 13, formerly 3A. That's where you find boron. Above it you will find the atomic number 5, and here you find 10.8. Yes, that value listed in the periodic table is precisely the average atomic mass, which is calculated by taking a weighted average for the element's isotopes.

[00:10:08]The percentages I showed you, 20% and 80%, are characteristic of the element boron. I can't expect the percentages to be the same for all elements, can I? That's something that will depend on the stability of that isotope. And then it varies from element to element. If I were to take, for example, chlorine, I would have different percentages for the chlorine. There would also be two isotopes, right, with percentages of approximately 25%, 75%.

[00:10:39]When I talk about oxygen, the percentages are different, so it depends on the element.

[00:10:45]Normally we don't need to spend that time calculating this weighted average because someone has already done it for us, right?

[00:10:54]So, we already have the values cataloged in the periodic table, and when we want the atomic mass of an element, we go there, look it up, see what the value is, and apply it, right? If you 're taking a test, depending on which entrance exam it is, you might have a periodic table with information on the test, you might have a data table, a list of data on the test, or it might even be within the questions you're solving that you need to use to apply the information, right?

[00:11:29]We'll see examples of this in a little while. Molecular mass, the sum of the atomic masses of the elements in a given form. I used aluminum sulfate, L2SO43, as an example.

[00:11:43]You see the atomic masses of oxygen, aluminum, and sulfur, right, which I wrote down below. And I want to know the molecular mass of aluminum sulfate.

[00:11:54]So, notice something. I'll start here with aluminum. Note that there are two aluminum atoms, therefore twice 27 units.

[00:12:04]I also have three times the amount of sulfur, 3 x 32 units.

[00:12:12]Now notice the oxygen: 4 x 3 equals 12.

[00:12:18]12 times 16 units.

[00:12:23]This is going to take me a while to calculate, right? 2 x 27 plus 96.

[00:12:33]And when you do the total sum, you get 342. It's not like I do mental math that fast, is it? I even left my calculator here so I wouldn't waste too much time with calculations and we could try to speed up our live class. But this molar mass, the molecular mass of aluminum sulfate, is a value that I use so often in class, you know, to solve problems, that we end up having some values ​​that teachers end up memorizing because of their use. But don't worry, you don't have to know this by heart, right? Because we calculate this as you just saw, we just have to be careful with the calculations so as not to mess up, right? But that gives us a molecular mass of 342 units.

[00:13:22]Well, that's a small amount, equivalent to 6.02 x 10 to the power of 23 particles.

[00:13:32]Normally rounded to 6 x 10^23.

[00:13:36]When I say particles, they can be atoms, they can be molecules, they can be ions, they can be electrons, they can be protons, they can be neutrons, whatever you can imagine, because this quantity, the mole, in the end, resembles things, like, for example, a dozen, which are obviously 10 things, right? A hundred, 100 things, a dozen, 12 things, right? So, the quantity mol is a fixed, fixed amount of things, let's say. So, if you say that a dozen is 12 things, 12 objects, 12 particles, then when you talk about 1 mole, you're referring to 6 x 10 to the power of 23 things.

[00:14:26]But that number is very, very, very large. Do you have any idea what the number 6 x 10 equals 23 might represent? Because look, knowing that it's the number six followed by 23 zeros, that's the easy part. The difficult thing, I think, is for us to grasp just how large, how high a power of 10 to 23 really is, you know?

[00:14:55]To give you an example, imagine you could count a mole of atoms by counting one atom per second.

[00:15:05]Imagine the time it would take you to count, counting one atom per second, you would count 6 x 10 to 23 atoms.

[00:15:14]Surely you would spend 6 x 10 at 23 seconds, right? But do you realize how long that time is? If you take that time of 6 x 10^23 seconds and convert it into hours, convert it into days, convert it into years, you find, you know, a time so absurdly large that it would be equivalent to more than 4 million times the size of planet Earth. And just so you know, our planet is estimated to be 4.5 billion years old, right? 4.5 x 10 to 9 years. So, to get an idea of what this order of magnitude from 10 to 23 is like, right? Because often we develop the habit of using a number, but we do n't always stop to think about its real meaning. If that number is too large, it gives me other information. It shows me how small a molecule can be, how small an atom can be, right? Okay, another example to give us an idea of ​​this order of magnitude of 10 to 23, imagine you were to take one mole of iron atoms. 6 x 10^23 atoms of iron. Each iron atom has an estimated size of very small, on the order of 10 to 10⁻¹² m. It's an extremely small thing, but if you take a mold of iron atoms and you imagine making a line, a line, you know?

[00:16:46]By placing one iron atom after another, you can make a row with 6 x 10^23 iron atoms. Do you have any idea how long this line is? The length of the line, folks, is greater than the distance from the Earth to the Sun.

[00:17:01]So that gives me an idea of ​​this absurd order of magnitude that is 10 to 23, right? Because that's what we find in terms of water molecules in 18 g of water. 18g of water seems like a tiny amount to our eyes, considering how many water molecules there are. This is what we find; it's the number of iron atoms we find in a 56 g piece of iron. That's the number of atoms there, right? Avogadro's constant.

[00:17:31]6 x 10 23 particles per mole approximately. It was aogadro, professor, who decided on this deal. No, it wasn't. It wasn't. The number is named in honor of the Italian Amedeo Avogadro, who had the idea that there would be a constant number for measuring the quantity of atoms, the quantity of molecules. He couldn't prove it. But then came the discoveries throughout the 19th century, the discovery of X-rays, then radioactivity, which made it possible for scientists from the late 19th to the early 20th century to determine the number of lawyers, right? And then, every now and then, if there's a little problem, we might need to use it. It's possible the problem will already give me the lawyer's number so I don't have to memorize it. Yes. Maybe not. Yes, it's also possible that the problem doesn't tell you the value, because that value is very well known, right? So we start from the premise that we need to know, we need to know how much it is. Suddenly a problem arises that I need to use.

[00:18:42]Molar mass, the mass of 1 mole, that is, the mass that corresponds to 6 x 10^23 things.

[00:18:50]expressed in grams per mole, numerically equal to the atomic mass. In the case of an element, for example, oxygen has an atomic mass of 16 units, and a molar mass of 16 g per mol. Well, then there's the atomic mass, in the case of a chemical element, or the molecular mass, in the case of a substance. I'll use H3P4 as an example.

[00:19:14]Look at him. I gave the atomic masses of the elements and I want to calculate the molar mass of H3P4, which will be 3 x 1. One is hydrogen plus 31 is phosphorus plus four oxygens, 4 x 16.

[00:19:32]And if there's one thing I always like to do, it 's to note the unit of measurement. It costs nothing. Sometimes we're too lazy to write down the unit of measurement, but it's good not to be, because when you get used to the unit of measurement, you reduce your chance of making a mistake. So, how much does that add up to? Well, do the math, you'll find 98 g per mole. The molar mass of phosphoric acid, H3PO4, means that in this mass of 98 g x 10^23 molecules of H3P4.

[00:20:11]Another quantity that will also be important in solving our problems is molar volume. Volume occupied by one mole, as the name suggests. In the case of ideal gases, this is where molar volume becomes extremely important for solving problems. Ideal gases under normal temperature and pressure conditions. And there is a difference between what these old, standard temperature and pressure conditions (STP) are and the modern ones. Older models considered a temperature of 0ºC and 1 ATM. But you might say, "But geez, how can 0ºC be a normal temperature?"

[00:20:55]I understand, a temperature of 0ºC is not very normal, but that's because the temperature chosen by default is the melting point of ice, and ice melts at 0ºC. That's how it was chosen as the standard.

[00:21:11]Molar volume determined experimentally.

[00:21:15]The molar volume was then found to be 22.4 L/mol.

[00:21:23]If I need this value to solve a problem, will the problem tell me?

[00:21:29]Not necessarily, guys.

[00:21:31]Honestly, it's best to memorize that little number, right? Memorize this number, you won't take any risks, okay? 22.4 L for each mole of gas. Note that it doesn't apply to liquids, and it doesn't apply to solids. Liquid and solid are completely different because they are more aggregated, have a smaller volume, and are denser, right? So this applies to gas.

[00:21:53]If it's under modern standard conditions (STP), there's a slight change in pressure.

[00:21:59]Instead of using one atmosphere, a pressure of one bar is used, which is slightly different because one atmosphere, if you convert it, is approximately 1.013 bar. It's almost the same, but note that the pressure is slightly different.

[00:22:22]This also makes the molar volume slightly different. And now it goes up to 22.7, okay? What if the problem tells me that the gas is at STP (Standard Temperature and Pressure), but doesn't tell me if it's old STP or modern STP? What am I supposed to do with my life, right? So, we use the old CNTP (standard, no-till) because, let's say, it's a matter of tradition, right? So, the habit of using these old normal conditions, then yes, if nothing is said, if the problem doesn't give any information, but you want to start in TP, then I recommend using those 22.4 L per mole, okay? It's that more common, traditional value. So, class, we've just seen the chemical quantities we'll need to move on to the next step, which is to perform stoichiometric calculations and determine the amount of reactants or products in a reaction. These quantities can be masses in grams, kilograms, or any other unit of mass. Well, it could be volume in liters, milliliters, whatever. Well, it could also be a number of moles, right?

[00:23:47]So, we're going to try to find out these quantities by establishing proportional relationships, which is nothing more than proportion. It's all about reason and proportion, all the time. Yeah, yeah, if you want to use the rule of three, go ahead. If you simply want to do a dimensional analysis, you can. The important thing is to always consider that the quantities involved in a chemical reaction are proportional. So we're going to use ratio and proportion all the time. I'm going to show you step by step, starting with this first example, how you should proceed. This is probably the simplest example I could choose for us to begin with, to get the hang of it and see how it's done. Look at that.

[00:24:33]The complete combustion reaction of propane is represented as follows.

[00:24:37]So there you have the reaction, right? C3H8, 5O2, giving 3CO2, 4H2O.

[00:24:45]Calculate the amount of substance that is synonymous with the number of moles of CO2 produced from 15 moles of C3H8.

[00:24:55]C3H8, which is propane. Well, what to do?

[00:24:59]Proportion, establishing proportional relationships between the substances involved, right?

[00:25:07]You have there, for example, a ratio in moles, notice the stoichiometric coefficients, 1 mole of C3H8 to 3 moles of CO2. I want to know the amount of matter in CO2. I recommend you do it this way. Just so you know, I'm going to start by writing N for CO2. What do I mean by that? I want to determine the number of moles of CO2, right? So, I represented it by N CO2. I want the CO2 mold number, okay? Okay, everyone, I'm going to write a fraction now.

[00:25:51]In this fraction, I will establish the proportional relationship between the amount of CO2 that the reaction produces and the amount of C3H8 that is present as a reactant. What is the relationship that 3 moles of CO2 are produced from, starting from how much? Starting from 1 mol.

[00:26:16]1 mol of C3H8.

[00:26:20]One thing I find important, and that I do every time, is to write down the chemical formula so I don't get confused.

[00:26:30]After writing that fraction, pay attention to the information, the data from the problem, okay? The real problem is those 15 moles, right? There are 15 moles of C3H8 there. I wanted to use the highlighter here. Ready. Here it is. 15 moles of C3H8. So what am I going to do? I'm going to get that C3 H8 mold number.

[00:26:54]There are 15. And I'm going to put it here multiplying 15 m of C3H8. In such a way that when I multiply the fraction by the amount of Mor in C3H8, I cancel it out. Having done that, my result is in Mor of CO2. The calculation is very simple, it'll come to 45, right? 3 x 15.45 mol of CO2, which will be our answer.

[00:27:20]Well, maybe, I don't know, but it's very likely that you were expecting me to use the rule of three. Yes, I just wrote the rule of three in a different way. Perhaps in this particular case, something like that won't make it easier for me to solve the problem, but depending on the situation, trust me. If you start doing it this way, which we call dimensional analysis, you increase your problem-solving capacity. Want to see how it would work using the rule of three? Let's go. Using the rule of three, we have 3 moles of CO2.

[00:28:03]3 moles of CO2 for every 1 mole of C3H8, according to the balanced equation. I want to know the number of moles of CO2 for every 15 moles of C3H8, which ends up being the same, right? 3 x 15 gives 45. Number of moles of CO2, 45 moles. It almost seems that the problem is solved more easily using the rule of three. In that case, I actually agree, it seems so, but when we go to problems where the number of terms, the number of factors given is greater, the use of dimensional analysis is much more efficient and you will see that you can solve the problems more quickly, with greater agility and speed, saving time, I think that's what everyone wants in the ENEM exam, isn't it? That stressful situation of trying to finish that exam on time. Good heavens, that's a problem. You don't have much time to do it, do you? So, with so little time to do it and the task being difficult, you go ahead and find a method that allows you to solve it more quickly, right? And that's just too good. So let's train this way.

[00:29:24]More people are arriving here. Good afternoon to those who are arriving. Let's go.

[00:29:30]Well, in problem two, I want to do a mass calculation. Notice that this is a neutralization reaction between sulfuric acid, which is H2CO4, and sodium hydroxide, which is NaOH. And I want the mass of sodium hydroxide, NaOH. I put the molar mass here in parentheses, right? 40 g per mole.

[00:29:49]Required to produce 355 g. For sodium sulfate, I have its molar mass here.

[00:29:57]Okay, I'm going to use the same technique I used before, but whose dough? The mass of the hydroxide.

[00:30:06]Our dimensional analysis, well, it requires only three steps, just three.

[00:30:16]First, identify what you want to calculate and write down what you want to calculate. According to the text, calculate the mass of NOH. Okay, so I'm already writing down the NOH dough. That's what I want. The second step is for you to write a fraction.

[00:30:34]In this fraction, you will put what you want to calculate in the numerator.

[00:30:40]Look, I want to calculate sodium hydroxide and the equation for moles of sodium hydroxide, given that its molar mass is 40 g per mole. So, twice 40 g of NaOH. If you want to write the number 80, you can, of course, feel free. Whatever.

[00:31:01]What do I have in the denominator? 1 mole of sodium sulfate and 1 mole of it, here's the molar mass 142, 142 g of sodium sulfate Na2SO4. Well, is that what I did? I established a ratio that I derived from the chemical equation, and this ratio tells me that 80 g of hydroxide, or 2 moles of sodium hydroxide, produce 1 mole of sodium sulfate. That's what I wrote here, right, in other ways of writing. But in the numerator, what I'm saying is that there are 2 moles of sodium hydroxide, and in the denominator, I'm saying there is 1 mole of sodium sulfate. In other words, it's as if I were saying that 2 moles of sodium hydroxide produce 1 mole of Na2SO4, right? So I have the mass of sodium sulfate that the problem gave me, which is 355.

[00:32:08]I'm going to multiply that by the mass of Na2SO4 and the problem will be solved with the mass of sodium hydroxide determined in the end.

[00:32:20]My dear friends, look, at this moment something happens that is very, very common in college entrance exams, which is the question writer purposefully choosing a number, a number chosen to make the calculations easier and make your life easier.

[00:32:41]And you can't afford to miss out on taking advantage of that. Why am I talking about this? Because if you take 142 plus half of 142, that is, 142 + 142, or in another way, 142 is not 71 x 2.

[00:33:05]I found an even better way than the way I was thinking. Look, where's my pen? Let me change the pen color here to make it stand out. Look, folks, this is nothing more than 2 x 71, right? This number here, look, 355.

[00:33:28]This is nothing more than 5 times 71. You see, this number 355 wasn't chosen by chance? Because when we think that way, we cancel 71 with 71, we cancel 2 with, and what's left? A very simple calculation that everyone knows how to do, which is 40 x 5. So the problem becomes very, very, very simplified. 40 x 5 equals 200 g of sodium hydroxide, right? And I don't know if you're there with your calculator in hand, right? Doing this calculation to check. And then you go and do all the calculations, it takes a lot of time. Sometimes using a calculator takes more time than doing it this way, which I did.

[00:34:28]Listen, when we're studying, preparing for college entrance exams, and we come across numerical questions like this one and so many others, well, it's something extremely important, you know? It's about keeping this object far away from us, so we're not tempted to use it, you know? It's about keeping your cell phone far away from you so you 're not tempted to use it. Because when you're in a hurry, in that eagerness to finish a numerical problem, you immediately pull out the calculator, grab your phone, do that little calculation, because you just want to finish that problem quickly and move on to the next one, right? So, you think you're saving time, huh? No, you're wasting your time.

[00:35:16]Why? Because you're missing the opportunity to look at the numerical problem, to find ways to simplify it mathematically, to find a proportion, a small number that helps you simplify, to round it off, to approximate it, I don't know, because you're using a calculator, you're using your cell phone and you won't notice that. And the training for chemistry, for physics, for mathematics, for those things where we, you know, encounter numerical problems.

[00:35:49]Preparation for college entrance exams needs to be like this. I need to look at the problem and find a way to finish it quickly, because I have very little time to work on a single question; the ENEM exam is such a hectic time, isn't it? So, follow this model and this method I'm using; it makes your life easier, it's better than the rule of three. Because when you use the rule of three, you probably won't notice fraction simplifications like I just did. And then you'll do one calculation, then another, a whole bunch of things, and spend more time than you should, okay? Follow the model, try it, and everything will turn out great. Let's see another one.

[00:36:34]Now, calculating the volume, I say that the thermal decomposition of dinitrogen pentoxide in the gas phase consists of the chemical equation 2N2O5 giving 4NO2 + O2 starting from 0.8 mol of N2O5, what is the total gaseous volume produced if measured at 0ºC and 1 ATM?

[00:36:55]0ºC atm. Under normal temperature and pressure conditions, the old ones, molar volume, 22.4 L for each mole of gas. The problem did n't say so, but we need to bring this knowledge to the table to solve this type of issue, right? So, look, I want to know what the total gaseous volume is. It's not the volume of NO2, it's not the volume of O2, it's the total. So, look at the total number of moles.

[00:37:26]I have five moles here that are produced, right? Since I want to know the volume, let's do it this way. Volume of what? Gas volume. That's what the problem calls for. I was advising you to start by writing a fraction and then, in the numerator of that fraction, writing what you want to find. What do I want, right? volume. So, five, five what? 5 moles. What is the volume of each mole? 22.4 L.

[00:38:02]It turns out that this amount of moles corresponds to 2 moles of N2O5.

[00:38:11]That's the amount of liters, right? 5 moles. 5 x 22.4 L is for 2 moles of the reagent N2O5.

[00:38:20]The text tells me that starting from 0.8 mol of N2O5, multiply, cancel, do the calculation and go up, right?

[00:38:34]So, what do I need to do now?

[00:38:35]Look how cool this is in terms of simplifying fractions, right?

[00:38:40]Escaping the calculator, escaping the cell phone, grab 0.8/2, 0.4.

[00:38:47]Take 5, multiply by 0.4. How much does that come to? That gives you 2.

[00:38:54]And then you take the 2 and multiply it by 22.4. And now it's become too easy.

[00:39:00]22.4 x 2, gas volume equals 44.8 L, right? I can do this calculation really quickly.

[00:39:12]I have my calculator here in my hand, just so you can see how sometimes we lose more time by using a calculator.

[00:39:18]Why would I take, like, 5 x 22.4 x 0.8 and then divide by 2, right? When I could just simplify the fraction and finish this calculation faster, right? And that's something that comes with habit, with training, this ability to perceive a simplification or perhaps even, if necessary, a rounding or something like that, right? Okay, let's get the hang of stoichiometric calculations here. I've already shown you how to calculate the number of moles, how to calculate mass, how to calculate volume, and now the number of molecules. For the number of molecules I need the number of lawyers.

[00:39:58]The decomposition of hydrogen peroxide contained in hydrogen peroxide solution is represented by the given chemical equation, right? Starting from 170 g of H2O2 given the molar mass 34 g per mol. How many molecules of gaseous oxygen are produced?

[00:40:16]Number of gaseous oxygen molecules.

[00:40:23]Let me see what's going on in the chat here. How did I calculate the volume? Let me go back a little to tell you. Oh, come back here. What did I do here in calculating this gas volume? The first thing I had to realize was that the problem was asking me for the total gas volume, in other words, what amount of gas I have in the product, right? I have NO2 and I have O2, with a total of 5 moles. Each mole under these conditions, 0ºC atm, occupies a volume of 22.4 L. The text mentions this, but where is that information? Where in the question does it tell me that 1 mole of gas occupies 22.4 L?

[00:41:11]The question isn't explicitly saying it, but if you come back with me, just hold on a little bit and we'll come back here. Where? Ready. Look, in item eight, at the very end of the page here, in item eight, the molar volume, the volume of 1 mol of an ideal gas at STP at 0ºC is 22.4 L/mol. So, what happens? There are problems that tell me this volume, there are problems that don't. So, I need to bring this knowledge to the table so I don't become dependent on whether or not the problem creator mentioned this specific volume to me. So that's 5 moles. Each mole equals 22.4 L. That's because I want to know the volume of gas. So, this amount of gas is produced by 2 moles of N2O5, which I put in the denominator, and multiplied by the amount of N2O5 given in the problem. So I did the calculations, simplifying fractions, you know, trying to find a way to finish quickly without, you know, going around in circles. And the volume came to 44.8 L, right? So I think a key point in this problem is knowing that one mole of gas at 0ºC atm, that is, at STP (standard temperature and pressure), occupies a volume of 22.4 L, right? That makes all the difference, doesn't it?

[00:42:39]Well, in this case four here, where I want to know the number of O2 molecules, I'll need Avogadro's number. Notice that it's 1 mole of O2 for every 2 moles of H2O2. So this ends up being a small fraction, right?

[00:43:06]Hey everyone, keep sending in your questions, okay?

[00:43:10]Isa Suzi also thanks you here.

[00:43:12]We're in this together. If you have any questions, let me know and I'll do my best to help us understand each other as clearly as possible.

[00:43:19]Well, in this chemical reaction, notice that 1 mol of O2 is in the product, 1 mol. And 1 mole has a number of molecules of 6 x 10 to 23.

[00:43:32]Well, it wasn't 6.02 or something. Yeah, but let's round it off, shall we? Because if you're thinking about it like that, oh, it's a multiple- choice question, right? So you don't need to be so strict when writing 6.02, right?

[00:43:46]Because the calculation will result in practically the same thing. So let's round it up to six and move on to the hug, right? So, folks, this here is the number of O2 molecules produced in this reaction. Is that correct? Since it's 1 mole, that's Avogadro's number. 1 mole corresponds to that amount.

[00:44:06]6 x 10 23 molecules, but that amount is for 2 moles of H2O2. So that's what I'm going to write down. 2 x 34 g of H2O2. Where did that 34 come from, kid? From here, look, 34 g per mole is the mass of 1 mole of H2O2. How much H2O2 did the problem give me?

[00:44:31]170 g. Let's multiply 170 g of H2O2.

[00:44:37]How much will it cost? Canceling a gram of H2O2, the number of O2 molecules is equal to what? Now all we have to do is a little calculation, a reasonably simple calculation, I would say, because if you look closely, see the simplification I'm going to do here. 34 is more than 2 x 17.

[00:45:01]If 17 cancels out here, look, with 170. How wonderful. Here's a 10.

[00:45:08]What's the score? There's what's left, look what's left. 6 2 x 2 4. 6/ 4. 6/ 4 is 3/ 2 is 1.5. Okay, the result is finalized. 1.5 x 10 to the power of 24. How did 24 appear?

[00:45:29]If it was 23, it's because it combined, look, 10 to 23.

[00:45:34]It combined with 10, 10 to 24. Number of O2 molecules, 1.5 x 10 to 24 oxygen molecules.

[00:45:46]Once again, something I told you a few minutes ago is happening. Whoever creates a college entrance exam question almost always chooses a number that makes your life easier.

[00:45:58]This number 170 here, folks, wasn't chosen by chance. That's because he was simplifying it with the 34 that was in the denominator, right? A sensible test creator does it this way. And that's usually how the group does it, right? I say this because you also create evidence, right? And then we think, "Wow, the student won't have a calculator handy to do this, so I need to choose a number that can make the calculations a little easier for them, right?" So, look, again, I'll repeat it for you, if you take your calculator and do this calculation, how much time do you waste doing this calculation, right? You're wasting precious time. Sometimes they even pick up their cell phone, they're studying, they pick up their cell phone, right? And then, when you open your phone to create an account, you see WhatsApp full of messages, you start browsing WhatsApp, and you get the urge to browse Instagram, and that's it, you don't study anymore.

[00:47:08]This is Sus. Let me tell you what I did here. Look, it was a multiplication, a multiplication, okay? 10 to 23 x 10. Look, repeat the basic steps, right?

[00:47:23]Add the exponents. That's why it ended up being 10 to 24, right? It must be because I did it too quickly. Then I said that I added 10 to 23 with 10, and that's when you got lost. But that's because I was multiplying 10 raised to the power of 2³ x 10.

[00:47:40]Then I repeated the base, base 10, and then added the exponents, 2³ + 1, right? Great, everything's fine, all good. Whatever, but I think so, it's a question, send it to me, it doesn't matter what it is, okay? Rest assured.

[00:47:57]Well, this is a slightly different, particular problem, isn't it? Perhaps it will require a little more from us, but who knows? We 're getting used to doing it, aren't we?

[00:48:11]Problem with reagent in access. Reaction between barium chloride and sodium nitrate in an aqueous solution as follows.

[00:48:20]Mixing 4.16 g of sodium chloride with 4.26 g of sodium sulfate. Which reagent is in excess? How many moles are in excess?

[00:48:32]First thing I'm going to do, class, and I recommend you do this too, is determine the number of moles, the number of moles of each one, so you know who is in excess, who has too much, who has more than necessary. Look, for sodium chloride, each 1 mole— see the molar mass—each 1 mole has a mass of 208 g. Even the amount of dough I put there, 4.16.

[00:49:03]This calculation is easy, because if it were 416 it would be 2. So 002.

[00:49:17]What is the number of moles of the other Na2SO4? Each 1 mol. Where is his molar mass? Here it is. 142. Each 1 mol weighs 142 g.

[00:49:28]Ah, where's his mass? It's there.

[00:49:32]4.26 g. How much does this cost?

[00:49:38]142 x 2 would give 200 and 84. I need to add a few extra points here, because look, the calculator is near me and I already want to use it, and we have to resist the temptation of having a calculator nearby, right? No.

[00:50:02]So, I was thinking like this, you see. 142 + 142, that already equals 284.

[00:50:11]And if I add another 142, that equals 62. Oh, how wonderful. I just discovered that 142 x 3 equals 426.

[00:50:25]Oh, how wonderful. Look how much this is!

[00:50:27]0.03.

[00:50:31]I just misspelled the word "mol," right?

[00:50:33]Nobody deserves that, right? Let's write this thing down properly.

[00:50:38]So, that's how it ended up, right? Right? 0.03 mol.

[00:50:43]And the calculator here next to me is begging: "Come here, use me, I'm here and I don't want to use this darn thing," right?

[00:50:51]So, we have to resist bravely.

[00:50:55]So the result is 0.03 mol. 142 x 3 426.

[00:51:04]Too much success. Look at that. So here's the thing. So how am I supposed to find out who's in excess? Look at the equation's balance. Note that 1 mole reacts with 1 mole. It's a one-to-one ratio there. So I'm realizing that this 0.03 here is too much. Because if the ratio is one of each, to react with 0.02 of the first, only 0.02 would be needed. You don't need 0.03 to exaggerate.

[00:51:35]Therefore, I conclude that of that 0.03, 0.02 reacts, leaving 0.01. Okay, so I'm already answering both questions here, because the first one is: which reagent is in excess? The second question is how many moles are in excess. So let's go. Yes, an excess agent.

[00:52:11]Excess reagent.

[00:52:13]It's sodium sulfate. Actually, if the problem were asking me what the limiting reagent is, which is the reagent that is not in excess, also called the minor reagent.

[00:52:32]So it would be BA Cl2, right? It would be BAC2 and the number of moles in excess.

[00:52:45]You just saw with me that what 's left is 0.01 mol. Therefore, I recommend that when dealing with problems involving excess reagent, you work with the number of moles, calculating the number of moles of the substances involved. The problem will be solved much faster; it's very practical to do it this way, okay? Look at the number of moles, and then you can develop it much more easily. Is everything alright with this one? Is everything alright with this one? Let me know if you have any questions. I'll pass this along here so we can take advantage of this topic. Problem: purity content.

[00:53:23]Iron production in the steel industry utilizes the chemical reaction shown in the following equation. Consider hematite ore containing 80% by mass of Fe2O3 and inert impurities. What is the mass of iron obtained from 400 kg of ore? I want to know the mass of the iron product? I'll give you the mass of the ore. And the first thing we need to do is determine the mass of Fe2O3 in this ore. Considering that the ore in question is 80% Fe2O3 by mass, nothing unusual about that. 80% of 400 kg, which is 320 kg.

[00:54:12]So the problem has two steps, because now I've just found out the quantity of reagent I'm going to use to produce this iron. And now I want to know what the mass of iron is. The stoichiometric ratio for balancing the equation is 2 moles of iron to 1 mole of Fe2O3 oxide, given that 1 mole of iron has a mass of 56 g, since there are 2 moles.

[00:54:41]2 x 56 g of iron. That's for one mole of Fe2O3.

[00:54:48]Where's his molar mass? 160 g.

[00:54:53]of Fe2O3. And how much do I want to use? 320 kg of Fe2O3. So I'm going to cancel the Fe2O3. But notice one thing, I'm canceling out grams with kilograms. Yeah, no, that doesn't make sense, right?

[00:55:10]I'm cancelling, folks, gram for gram, but I haven't canceled the kilo prefix. 1 kg equals 1000 g. So, leave it alone, leave 1 kg. If you want to convert everything to grass, feel free, multiply by 1000, no problem at all. But if you want to leave it in kilograms, that works too.

[00:55:33]Look how easy the calculation became. 320 by 160 2, right? So what was left? 2 x 56 112 multiplied by 2 224. mass of iron obtained 224 kg. And that? Hey, how's it going? In this little problem involving purity levels, where it gives me the mass of the ore, 400 kg of ore, iron ore containing Fe2O3.

[00:56:06]It has 80% purity. So I immediately applied the 80% to the 400 kg to understand that in that ore, in those 400 kg, I have 320 kg of Fe2O3.

[00:56:20]And then I used the reaction ratio of 2 moles of iron to 1 mole of Fe2O3 to find out the amount of iron that the problem asked for, right? 320 plus 160, I simplified it to 2 x 56 x 2, resulting in 224 kg. I'll pass this along so we can look at another case.

[00:56:45]Problem involving yield.

[00:56:48]The catalyzed combustion reaction of hydrazine occurs as follows.

[00:56:52]Hydrazine is N2H4.

[00:56:56]How many moles of NO are produced from 128 g of hydrazine with a molar mass of 32 g/mol? Considering a return of 75%.

[00:57:09]Well, look at that, the number of moles of NO.

[00:57:15]And I'm already looking at the equation, realizing that it's 1 mole of hydrazine for every 2 moles of NO.

[00:57:23]I want to know the number of moles of NO. I'll start by putting him here in the numerator. 2 moles of NO for how much hydrazine? 1 mol. How many grams is 1 more mole? 32 g of N2H4. Notice the molar mass, right? 1 mol equals 32 g. And I want it starting from 128 g of N2H4 hydrazine. plus an additional return is 75%.

[00:57:52]So I'm going to multiply this by the 75% yield. Okay?

[00:57:59]Why? Because of the reaction with the 75% yield, what does it imply to me? The thing is, I won't be able to get everything I could. So, if I don't write 75%, I'm implying that I produced 100%, that I produced everything, right? But if the return is 75%, it's because there was a loss, and I'll only be able to recover 75% of the total.

[00:58:25]So I take that amount and multiply it by 75%.

[00:58:30]Okay, so I'll give you a tip for working with percentages like this: notice that 75% is nothing more than 3/4. This helps a lot with a problem like this. Want to see? 128 divided by 32, what do you get, boy? 128 divided by 32 equals 4. How wonderful! 128 divided by 32 equals 4.

[00:58:59]Take that 4 and cancel it out with this 4 here. So what happens then? 3. And you're going to multiply by 2 here. And the result will be 6.

[00:59:11]Conclusion, the number of moles of NO is equal to 6 moles.

[00:59:19]So you can see once again how dimensional analysis, which is the technique I'm using and not the rule of three, can make your life much easier. Because when you organize, when you structure it in this way that I do, like a compound rule of three, everything becomes easier. You save time, you simplify things, you find a fraction there to replace a percentage, right? A number appears in the numerator that cancels out with the number in the denominator, and then the problem becomes much more straightforward.

[00:59:52]Considering all of this, we have another case to look at here. Come with me.

[01:00:02]Production, the industrial production of sulfuric acid, is done using this sequence of steps.

[01:00:09]What is the maximum mass of H2SO4, 98 g per mole, that can be obtained from 3.2 kg of food-grade sulfur?

[01:00:21]When he says food-grade sulfur, I'm talking about S8 here.

[01:00:25]Elemental sulfur.

[01:00:28]Atomic mass of sulfur, 32 units.

[01:00:32]So, here's what I'm going to do in this one.

[01:00:35]Well, one important thing to realize is that in a problem like this, I would have to obtain a global equation by summing the steps, right? So, for that to happen, I would repeat the first one, but for the second one I would have to multiply by 4 to get 8 SO2 here.

[01:00:58]Look, look, multiplied by 4 to cancel SO2.

[01:01:04]And the last one multiplied by 8 would look like this, multiplied by 8 so I can cancel SO3, you understand? So when I add those three steps together, I would have S8 + 12 O2 + 8H2O.

[01:01:29]And here in the product I would have 8H2SO4.

[01:01:34]And then I would realize, through this sum of steps, that looking at the overall equation I arrived at, I would realize that 1 mol of S8 produces 8 mols of H2SO4, right? That's what we're seeing from the overall equation.

[01:02:01]Although, honestly, I wouldn't need to spend all this time here, look, to realize that one molecule of S8 should produce 8 of H2SO4, that 1 mol of S8 would produce 8 of H2SO4, because after all, S8 has eight sulfur atoms, so surely I would get eight of H2SO4, right?

[01:02:26]Okay, now that that's done, I want to know the mass of H2SO4 that I would be able to obtain. So, what would be the mass of H2SO4 equal to 8 x 98, which is the molar mass of H2SO4, right? And here in the denominator I need the molar mass of S8. It would be 8 times 32 g, because, you see, the atomic mass of sulfur is 32, but this is just one sulfur molecule.

[01:03:02]Since I'm thinking in terms of S8, I would need to multiply by 8. This multiplied by 3.2 kg of S8.

[01:03:14]This calculation ended up being relatively simple after all that, don't you think?

[01:03:20]Look what happened here!

[01:03:22]Canceled 8 with 8. 3.2 with 32 equals 0.1.

[01:03:27]Multiplying that by 98 gives you 9.8 kg.

[01:03:33]So you've just seen how a little problem involving reaction sequences works, and the importance of realizing that the steps being added together involve the cancellation of some substances— one that is produced in one step is consumed in another—to arrive at a global equation. This isn't always necessary, in fact, because, for example, in this case, it would have been enough to realize that if S8 has eight sulfur atoms, one mole of S8 would produce 8 moles of H2SO4, without needing to sum the steps to figure that out.

[01:04:07]But at least the sum of the steps gives us a better understanding of what would be done.

[01:04:16]We have therefore seen various forms of stoichiometric calculations, in diverse situations, right? To recap what you've seen, it's possible to calculate the number of moles, the mass, the volume, and the number of molecules.

[01:04:31]We may have a problem involving excess reagent, purity level, yield, or a problem resulting from reactions.

[01:04:40]Time to apply, keeping in mind the college entrance exam problems, okay? Well, from the ENEM exam or in the ENEM format, so you can see how you're going to apply everything you just learned here with me, right? such as, for example, this commercial product known as liquid limestone. It is a mixture used to reduce soil acidity, which contains approximately 17.5% calcium by mass.

[01:05:09]Given the molar mass of calcium, 40 g per mol, the amount of calcium ions in moles present in 1 kg of the product is approximately: So, let's see. 1 kg of the product. The text then says, in the first or second line, that this liquid limestone is a mixture used to reduce soil acidity and that it contains about 17.5% calcium by mass. So, I can now apply that 17.5% to the total mass to find out the mass of calcium.

[01:05:47]It stays like this.

[01:05:53]17.5% being 1 kg, let's put 1000 g.

[01:06:00]So I'll get 100g and 75g, right? Yes. 17.5%, right? If it's 1000 g because it's 1 kg, then 175 g.

[01:06:15]And I want the quantity of calcium in moles.

[01:06:20]number of moles of calcium.

[01:06:25]So you saw that the molar mass of calcium was given in the third line of the problem, right? Each 1 mol weighs 40 g.

[01:06:34]We have 175 g to find the number of moles.

[01:06:41]Is it possible for us to make a reasonable approximation here?

[01:06:45]If instead of 175 it was 160, then 160 would give exactly 4. So, do you know what that is? It's a number slightly above four. 4.

[01:07:04]Mols.

[01:07:05]Look what I came up with here to try a quick and straightforward calculation. The thing is, if instead of 175, if instead of 175 here it was 160 g, I would divide 160 by 40, the result would be 4, right? But since the number is 175, which is slightly greater than 160, then my result is slightly greater than 4.

[01:07:33]And what I have in the alternatives that is slightly greater than 4 is 4.4, right? If you try to calculate it precisely, it's a waste of time. Once again, 175 divided by 40 equals 4,375. There's no need for that, right? You make an approximation there, you know the result will be close to the number four, but then you make a small adjustment, right? A little more than four. In that case, 4.4, right?

[01:08:03]Question number two. Brazil is the world's largest producer of niobium with a molar mass of 93 g per mol, a metal used in the manufacture of various types of automotive, structural, and stainless steels. The process used in the production of niobium is the aluminothermic reduction of NB2O5 with a 10% excess of aluminum, which has a molar mass of 27 g per mol relative to the stoichiometric quantity of the reaction represented by the given chemical equation. A metallurgical engineer estimated the mass of aluminum needed to produce 9.3 kg of niobium under the described conditions for the production of a batch of steel parts ordered by an industry, considering a 100% yield.

[01:08:51]The mass of aluminum in kilograms estimated by the engineer is closer to "didn't understand the problem properly." Look, what I need to find out is the mass of aluminum needed to produce a certain mass of niobium, right? Molar mass of aluminum: 27 g per mole.

[01:09:12]Molar mass of niobium: 93 g per mole. And there's another important piece of information here in the text, on the fourth line. When it says there is a 10% excess of aluminum in relation to the stoichiometric amount, that is, it means that I need an additional 10% to determine the amount of aluminum. So let's see here. I want to produce 9.3 kg of nickel. Let's see what the mass of aluminum is in that area.

[01:09:47]Oh, look at the equation over there. Somewhere in the middle of the text, the chemical equation is given: 10 moles of aluminum to 6 moles of ion. I'm only going to highlight this part, okay? 10 aluminum, producing six from niobium. Molar mass of aluminum: 27 g per mole. Molar mass of niob 93 g per mol. Ah, the amount of nickel I want to obtain is 9.3 kg. I want to know the mass of the aluminum, given that it has a 10% excess. Ready. This is a brief summary of the text from the previous page. When calculating the mass of aluminum, I always like to note down which substance I'm referring to, so I do n't get confused.

[01:10:37]I want the aluminum paste. There are 10 moles.

[01:10:39]Look at the coefficient. 10 x 27 g of aluminum. That's to produce 6 tons of niobium, 6 times 93 g of niobium, and I want to produce 9.3 kg of niobium, but I have to add, as the text tells me, an excess of 10%. An excess of 10% means that it is a mass 10% higher than that mass, or in other words, it is a mass 0.1 higher than that mass.

[01:11:18]Wow, that "M" here came out really weird. Let's sort this thing out.

[01:11:23]So, what is this? That's 1.1 times that mass. Then multiply by 1.1 and subtract the mass of the aluminum from that.

[01:11:34]All done. Okay, let's look at some simplifications, rounding, and things like that, so we can, you know, do this faster. Look, first, this one multiplied by this one equals 0.1.

[01:11:48]10 m by 0.1 is enough, that's one, right? No need anymore. So, the idea is to take that 27, divide it by 6, and multiply by 1.1, right? Is that an exact division?

[01:12:02]27 by 6.

[01:12:06]27 by 6 is what I have to do.

[01:12:09]What is this? 3 x 9.

[01:12:13]3 x 9 divided by 6. So, cancel here, it gives 1 over 2, which gives 4.5, right? Hey, you're on the right track, right?

[01:12:23]4.5, but I have to add 10%, so 4.5 + 0.45, which will give almost 5.

[01:12:37]Mark the letter E, approximately 5 kg. Look, the question itself implies that you don't need to get the exact result, right? He's asking for an approximate result regarding the mass of aluminum you'll need to make this niobium. So, if you do the exact calculation, it comes to 4.95. There's no need for that, right? Then you make an adjustment, a rounding, a simplification of the fraction, and it ends up working out, which is what we really need to do in a test like the ENEM, right? A way to save time and take care of other things.

[01:13:17]Well, a sedan typically features 200 kg of aluminum distributed throughout the chassis, engine, and cabin. A sample of bauchite, the main natural source of the metal, is composed of 50% aluminum oxide by mass. Consider the molar mass of aluminum to be 27 g per mole of oxygen to be 16.

[01:13:38]The mass of balchite, which must be used to produce the aluminum used in the manufacture of a car of this model, is closest to it. A sedan car typically contains 200 kg of aluminum.

[01:13:53]Well, then he says more: a sample of bachite, the main source of the metal, is 50% aluminum oxide by mass.

[01:14:03]OK? And he asks what mass of bachite should be used to produce the aluminum used in the manufacture of a car. OK? So here's the thing, the compound is Al2O3.

[01:14:20]Let's find out what the mass of Al2O3 is.

[01:14:24]I will need the molar mass of L2O3.

[01:14:30]I want to know the mass of bachita. I have the molar mass of aluminum, right?

[01:14:35]Well, that's 200 kg of aluminum. Okay, so let's get to the L2O3 story.

[01:14:47]Okay, I'm going to do a quick calculation here to see what the molar mass of Al2O3 is, okay? The molar mass of L2O3 is 2 x 27 + 3 x 16.

[01:15:04]This will give me its molar mass in grams per mole.

[01:15:10]Molar mass of Al2O3 = a. That makes 2 x 27.

[01:15:16]54 plus 48. That gives 102 g per mole, right? The calculation is 54.482 g/mol. So here's the thing, for every 102 g of Al2O3, notice that it has twice the amount of aluminum. It has twice the amount of aluminum because Al 2. So I will have twice 27 g of aluminum, right? But I want 200g, 200kg of aluminum.

[01:15:55]Well, before doing the calculations, this would give me the amount of Al2O3 I would need to obtain 200 kg of aluminum. The problem asks for the mass of balchite, knowing that it is 50% aluminum oxide by mass. So, if I figure out how much this is, I take double the amount and the problem is solved, right? So let's see how we can simplify things.

[01:16:24]First of all, 200 divided by 2 equals 100.

[01:16:28]Here it equals 100, right?

[01:16:32]Another thing we can do is 102. Look, a reasonable approximation we can make is 102/27. If instead of 102 this was 100 and instead of 27 this was 25, this would give a ratio of 4.

[01:16:54]It would give a ratio of 4. So I'm realizing that this number would give something close to 400.

[01:17:04]Something close to 400, right? 400 kg of L2O3.

[01:17:13]It turns out that this L2O3 only represents 50% of the total. So, what exactly is the mass of a bauchita? The total mass. These 400 kg do not represent the total, but rather 50%.

[01:17:32]And I want to know what 100% is equivalent to. Actually, it's just doubling the value, right? And then I would get 800 kg. Actually, it has to be a little smaller, because when I rounded it here, I went from 27 to 25, so I reduced the number, right?

[01:17:51]And the result I obtained is above the expected value. But then the value I'll get should be a number close to 800 and less than 800. You can see there's a number 756 there, right? That 's a number less than 800, but it's close to 800.

[01:18:09]102 x 200 divided by 54.

[01:18:13]This times 2, see, once again rounding, useful approximations to arrive at the answer, right? OK?

[01:18:24]So, I'll do the calculation here just to make sure you do what I just showed you, because when you do it on the calculator, you get 755.55.

[01:18:39]And this is just to show you that there's no need to worry so much about calculations when you're working with approximations, rounding, simplifying fractions, and things will just flow smoothly.

[01:18:55]Four. Ethanol, a fuel widely used for energy supply, is obtained from sugarcane through the fermentation of sucrose according to the given equation. The density of ethanol, molar mass, and molar mass of sucrose are given. A fermentation process was carried out using 85 kg of biomass, which contains 45 kg of sucrose.

[01:19:19]In the end, the ethanol production yield was 85%, and the amount of ethanol obtained in liters is closest to that figure. With an 85% yield, he wants to know the quantity of ethanol in liters.

[01:19:34]So, let's come back here.

[01:19:40]There is a molar mass in this mixture; I used 45 kg.

[01:19:45]Well, the reaction shows me that the ratio is 1 mol to 4, right? So, I have 4 moles of ethanol, each mole... So here I have the density because I want to know the volume, okay? So, I'm going over here to do it this way.

[01:20:02]I want to know the volume of ethanol C2H5OH.

[01:20:10]I'll start with density. Each 1 ml of ethanol C2H5OH corresponds to a mass of 0.8 g. This is the information I got from the density, because if the density is 0.8 g/ml, then that means that each 1 ml has a mass of 0.8 g. From the reaction, I have that 4 x 46 g of ethanol is the mass that corresponds to 342 g of sucrose C2H22O11.

[01:20:51]I have 45 kg of sucrose, 45,000 g of C12 H2O11, and there's still a small yield of 85%, which I multiply by 0.85.

[01:21:07]And what a tricky little math problem that was, huh? Isn't that right? Look, this one was really hard work, canceling out grams of ethanol, grams of sucrose, right, to find that volume, it would be in milliliters, but since there's a 1000 here in the 45,000, then the result would actually be in liters, right? The result would be in liters. And that. So then you just do the math, do your rounding, approximation and all that, right? To unravel this little problem. For example, you can simplify 4 with 0.8.

[01:21:54]4 plus 8 equals 5. Another thing you can do with the numbers I'm seeing here, uh, this 46 can also be simplified to 342 a little bit. I know it's possible to make a quick little adjustment there, right? And then you'll be able to solve the problem and find your answer. The problem is that, in this case, the rounding is a bit tricky because we have numbers like 24 and 26, which are very close numbers, right? So, look, this here, if we calculate the exact value, this comes to 25.7.

[01:22:39]So, look, whoever creates this kind of problem really wants you to do a lot of calculations, because, you know? And this is from the ENEM exam. And in this case, a lot of calculations were needed to find the right answer so you could end up here with a number close to 26, a number very close to 26 L.

[01:23:03]In fact, at the very end here is the answer key for all these questions for reference, so we can reinforce our understanding.

[01:23:16]And I'm going to show you the last one here to give you an idea, because we're almost at the end of our time, right? But look, an industrial tank containing chlorite acid ruptured, leaking 365 kg of this substance.

[01:23:32]The accident containment team was mobilized and used hydrated lime to neutralize the acid. To avoid leaving any acid residue as a safety margin, 50% more of this base was applied. The neutralization equation is represented as follows. The molar masses are given as 36.5 and 74.

[01:23:53]Considering it to be 100% pure hydrated, the mass of C H2 used in the neutralization of the acid was... So you have a quantity of 365 g/kg of hydrochloric acid.

[01:24:11]You have 2 moles for every one here. You have the molar masses, and then you find the mass of COH2, which is what the problem is asking for. What does one mole of it represent in the balanced equation? Where is his molar mass? 74 g. Right at the very end of page 74.

[01:24:34]So 74 g for 2 moles of HCl times 36.5 g.

[01:24:45]So, at the beginning of the problem, second line, second line of the problem, 365 kg, 365 kg of HCl, right? But there's a small detail here. Let me see, there's some more information. When it says, in order to leave no acid residue as a safety margin, 50% more of this base was applied. So, if I want to know the amount of base that was used, I'm going to take this amount that I'm calculating here and add 50%, that is, I'm going to multiply it by 1.5, which is to add 50%, right? Cancel HCL, this one for this one adds up to 10, right? So I'm left with 74 times 2, which equals 37. I'll adjust this from here. That makes 370.

[01:25:38]If I take 370 and add 50%, then adding 50% here will give what? It will weigh 55 kg, right? So that's it. 370 multiplied by 1.5, that is, 74 x 365 divided by 1, meaning multiplied by 1.5 divided by 2 divided by divided by 36.5, which gives 555, as I was already projecting and checking here, just to avoid any problems, right? 555 kg here. So that's it, folks. Look, you're realizing how much work this subject can be, right? It can be a lot of work for you, because it requires us to have the patience to do the calculations carefully, right? You need to be familiar with the various cases, you need to practice a lot, right? But that's it, with training, I 'm sure you'll do well in stoichiometric calculations, and also knowing that this is an indispensable subject for learning other subjects as well, okay? Just a reminder that the second list of questions is available until Saturday, May 9th, right? You can access it through superenem.com.br, and it's available to anyone who has registered on the platform. I hope this helped you guys again, okay? Just one more of these classes here, and keep studying, keep studying, there's so much to learn, and you can count on us, you can count on Super Enem in this journey. A hug.

[01:27:28]Thank you all.