Unit Conversion & Stoichiometry | Important Formula | Easy Concept | JE-Chemical | HRRL

[00:00:00]Hi everyone welcome to my channel. So kids, this is the last chapter of your syllabus which was pending, Unit Conversion and Stheiometry. Let us finish this chapter today in this one video lecture. There is nothing much in it. Look, there is no point in just memorizing the units and then just keep studying. Isn't it? Regarding this, let's say you should know the basic unit. Let us discuss for a while what is basic convergence? What is stichometry? Let's discuss it a little bit. Whatever else is left, we will practice questions about its numericals and where they will be used.

[00:00:35]Ok? Your main command over it will be created there on the questions.

[00:00:41]Once you see the unit convert and steriometry. Let's discuss it a little bit. I will finish it in just this one video lecture. Ok? Sit down with a copy and pen. You can take note.

[00:00:54]As this is a length this unit converts. Unit conversion like meter and centimeter, we say that there are 100 cm in 1 meter.

[00:01:01]Right? So that's what converting units is all about.

[00:01:07]Ok? So if we talk about length. When we talk about length, we take it in meters. Take it in kilometers. So you know 1 km. What is it equal to? Equivalent to 1000 meters. Ok? And what is 1 meter equal to?

[00:01:29]Equal to 100 centimeters. In this way, if we talk about area, then the area is taken in square meters. Mostly you will see it shown in meter square and centimeter square.

[00:01:39]Ok? If you want to convert 1 meter square, how much will it be?

[00:01:45]10000 cm Equal to square. Right? In this way all these units can be converted.

[00:01:53]If we talk in terms of volume, then we take volume in litres, in ml, if we talk about volume, then if we talk about 1 litre, then it will be equal to 1000 ml and 1 metre will be equal to 1000 litres, okay, if we talk about mass, then the main unit of mass is 1 kilogram, mass is taken in kilograms, the relation with which will be 1 kg in grams. = 1000 grams and 1 tonne is what is 1 tonne equal to? 1000 kg.

[00:02:37]clear? is it done. Let's talk about pressure.

[00:02:40]Now about the pressure.

[00:02:43]What is the unit of pressure? One atmospheric pressure. Whose equal will it be?

[00:02:52]Equal to 1.01 times.

[00:02:54]Ok? This one atmospheric is also equal to 3760 mm of mercury.

[00:03:02]You just have to remember this much.

[00:03:04]Ok? There is no need to do anything too deeply. I will get the rest of the questions done for you.

[00:03:09]You have to practice. Ok?

[00:03:11]Talking about energy, there is a conversion for energy, like if we take it in calories, then 1 calorie will be 1 484 joules, okay, you have to remember that your temperature will be given to you in degree Celsius, if you want to convert it into Kelvin, then what you will do is add 273 to whatever is the value of degree Celsius, your answer will be in Kelvin. If you have a relation between Fahrenheit and Celsius, then the value of Fahrenheit = 9.5°C will be + 32. Here these two are the two ways to convert degree Celsius to Kelvin and you can convert degree Celsius to Fahrenheit or Fahrenheit to degree Celsius in this way.

[00:04:06]Ok? That's two for the temperature.

[00:04:15]After this, if we talk about this unit converge stomatology. If we write any reaction a + b c + d then what are these a b d? are the stoichiometric coefficients. Ok?

[00:04:35]What is this? All of them are stoichiometric coefficients. So, we have to check this after properly balancing the reaction.

[00:04:45]It provides very important information.

[00:04:49]Which reactant is reacting with which in what quantity, how much product is being formed, conversion, yield, how much amount is required to react A with B. So we get to know all this from the stheiometry coefficient only. So stomatology means it is very important. It is very important.

[00:05:08]Ok? If we want to discuss any parameter of any reaction, then first of all we have to start with stomatology.

[00:05:14]Now if we look at it regarding the reactions, one is the child mole concept.

[00:05:20]You know the concept of mole, right? One mole is how many atoms, particles are there in it? You can take 6.022 or 23. Multiply 10 to the power 23 particles.

[00:05:34]Ok? And what else is this value called? Agodro number. And what is it called?

[00:05:41]Agodros number.

[00:05:46]Ok?

[00:05:49]After this we look at the molecular weight. So you must know how to calculate molecular weight. Do you know the molecular weight? What happens?

[00:05:57]Sum of atomic masses is the mass of all the atoms, add it up. For example, if we want to find the molecular weight of water, then the mass of one hydrogen is one. So here there are two hydrogens, so multiply it by one.

[00:06:12]Plus one oxygen has a mass of 16. So here there is one oxygen, so its mass becomes 16. So what is the total? 18.

[00:06:21]Ok? So who got 18 grams?

[00:06:23]Molecular weight of water. So in this way, add the atomic mass of all the atoms plus their number, multiply it, add it with that and from here you can find out in this way, molecular weight is one, children, number of moles, number of moles is equal to what, mass over molar mass, mass over molar mass, okay, so from here we can find out the number of moles.

[00:06:55]For example, we can take that you have been given 36 grams of water. Ok? Its moles have to be removed.

[00:07:07]So what is the formula for finding moles? Mass. This is the mass of water.

[00:07:11]So 36 will come up. Just calculated the molar mass.

[00:07:14]How much do we have? 18. So divide this with water. So two moles will come. Ok? So this way we can find the mass number of moles from this fuel.

[00:07:25]So what is a balanced chemical equation?

[00:07:31]So if there is any reaction like A A B C and D. Ok? If it is properly balanced.

[00:07:44]Ok? Where will we find out from? A & B, C & D were its stoichiometric coefficients. Along with this we have many laws like conservation of mass, energy, all these are different laws. He will try applying.

[00:07:58]Which reaction is based on Agodros Law, Galzach's Law? Is it in liquid fuzz or solid fuzz? What are the components? Ok? Is the reaction homophasic or heterophase? We discuss every single thing inside this. So these are all the reactions, to discuss any parameter of any reaction, it is very important to balance the reaction first. If you have reaction balance, whenever you get any numerical question, that is, in the exam, first of all you have to see whether you have given reaction balance, if not then first of all balance it.

[00:08:31]Only after that, if you want to calculate any parameter, you can calculate it.

[00:08:36]Ok? This is a stoichiometric mass calculation, so you will choose stoichiometry here.

[00:08:41]Only after that you will use them further in calculations. One parameter is the mole fraction.

[00:08:51]Mole Fraction. Ok? The mole fraction of any component of any substance has to be calculated.

[00:08:56]Suppose there are two components in a mixture a + b. Ok? So you have to find the mole fraction. So what will you do?

[00:09:04]Whatever number of moles there are of that component, you will divide it by the total number of moles.

[00:09:12]Ok? That will come to mole fraction. So if there is A and there is B then the number of moles of A is NA and the number of moles of B is NB. So if we want to find the mole fraction of A, then the number of moles of that A i.e. and divided by the total number of moles. And if we want to find the mole fraction of B, what will it be?

[00:09:31]Number of moles of B divided by total number of moles.

[00:09:37]Is it clear?

[00:10:01]With this, now we do molarity and molality. You can also note down the formulas of molarity and molality.

[00:10:18]So now let's look at molarity next.

[00:10:27]Molarity. What is molarity? Number of moles of solute dissolved in 1 liter of solution.

[00:10:33]Number of moles of solute dissolved in 1 liter of solution. And now look here, who all are in a relationship? doing molarity. We have r here is the relation between solute and solution.

[00:10:53]Remember what is inside molarity? If we look at the formula of molarity, we represent it with a capital m. So m = number of moles of solute that have dissolved in the solution and what will be the volume of the solution in litres, so from this we can calculate the molarity. Now one more thing kids is the number of moles of solute above.

[00:11:22]So if we want to do this in mass, then we know what the number of moles is equal to?

[00:11:26]Mass over molar mass. So we can find the value of number of mass moles of solute here. So what will be the number of? Mass of solute.

[00:11:41]Ok? Instead of the number of moles, multiply the molar mass of the solute by one over the volume of the solution in a liter.

[00:11:56]Ok? So this becomes molarity.

[00:12:00]After this, you should know one more term. I am getting only as much done as is necessary for your exam. Ok? Molality.

[00:12:10]What is molality? Number of moles.

[00:12:16]Number of moles of solute dissolved in kilograms of solvent.

[00:12:22]This is represented by molality with a small m. Ok? Now in this the solute will have a relation with the solvent. Ok?

[00:12:31]So what?

[00:12:33]Number of moles of solute dissolved in 1 kg of solvent. Ok? I am getting it done for only two terms.

[00:12:48]Molarity and molarity. I am not getting normality done. You have a new need for it.

[00:12:53]Ok? So whose equal will it be? m = moles of solute If you want to make a formula, moles of solute divided by volume of solvent is fine, not volume, mass of solvent is fine.

[00:13:16]In what will the mass of solvent come, if it is in kilograms then this will become molality, kids, is it fine? So if you want to convert kilograms into grams below, you can convert them into grams.

[00:13:30]Multiply it by 1000. clear? So this comes to molality.

[00:13:38]Further we read that there is a formula called ppm. There is a term called ppm.

[00:13:43]PPM means parts per million. Ok?

[00:13:47]Like if there is 1 liter of water, there is 1 liter of water.

[00:13:52]1 mg of anything dissolved inside it.

[00:13:54]So if 1 mg of anything is dissolved in 1 litre of solvent or 1 litre of water, then it means it is equal to 1 ppm. Ok? If 2 mg of something is dissolved in 1 liter then it is equivalent to 2 ppm. Is this clear, child? So here we are talking about ppm i.e. 10 parts per million i.e. 10 to the power 6 million. So if we want to calculate ppm from here, then the formula for ppm is mass of solute divided by mass of solution * 10 to the power 6, you can clearly note this, parts in million.

[00:14:50]After this, children, if we talk about gas, if we are talking about the volume of any gas, we are giving the condition STP standard temperature and pressure. At STP, if we talk about the volume of gas, then one mole of gas is equal to 22.4 liters. It will not give it to you, it is assumed that you have read it.

[00:15:14]Friends, you will see this question commonly in which you will have to convert it into liters. You should know that 22.4 liters is STP condition. It will give you special writing. Okay, so what is equal to 1 mole of any gas is equal to 22.4 liters. You should know this.

[00:15:32]After this, let us read a term, limiting reagent.

[00:15:43]Look inside this, children, there is a limiting reagent. There is a limiting reagent. There is an access reagent.

[00:15:52]Access reagent is a for Example 1A + 3B react with each other and give a product CD. Let us have nothing to do with the product. In this we have reagents which you will know about access and limiting from the reactants. Ok? Will not give you difficult level questions. Even if he gives whatever type of question comes, after this we are going to start the revision section, revision and previous year questions, so we will solve everything. Ok? There is nothing to worry about. So limiting reagent like this is suppose to be a balanced reaction.

[00:16:30]This is a balanced chemical equation. It is balanced. Now what is in the question?

[00:16:37]In the question he said take one mole of A reacts with 2 moles of B like this. 1 mole of A reacts with 2 moles of B, right?

[00:17:03]Or don't react like this, meaning okay. He is making me react like this. Given that 1 mole of A reacts with 2 moles of B, let's find out which is the limiting reagent?

[00:17:13]Which is the access reagent? Now look at this, children, first of all you wrote the chemical reaction of whatever A and B would be.

[00:17:21]Checked the balance.

[00:17:24]By balancing, you found out that 1 mole of A is required and how many moles of B are required? 3 moles are required. Now we saw how many he is providing inside the question. Only one is being provided for question A.

[00:17:35]Ok? But B is giving him two.

[00:17:38]How many do we need? I need three.

[00:17:41]B He is giving one price less. What does this mean B? Limiting reagent here.

[00:17:49]is the limiting reagent. It will be over before he reacts with A. Is that okay?

[00:17:54]As much as it is because I wanted more, I wanted three. But if it is less than three then it will be finished before the reaction is completed.

[00:18:02]So what is this? What is B?

[00:18:05]is the limiting reagent. Ok? But if he had asked the question here in a slightly different way, that 1 mole of A reacts with four moles of B, then this time A is giving only one. But B he gave four moles. We should have had three.

[00:18:22]But I gave one extra price for B. So A and B will react to give 3 moles of B and back still one B will be left pending which will get wasted inside the reaction. So in that case we'll say now what is our B this time? Access Reagent. Do you understand? This time our B is access reagent. So this is how one has to look inside the reaction.

[00:18:46]Which is the limiting reagent? Which is the access reagent? There are a lot of numericals involved in this but we will look into it if you need it, there is no need to do it right now.

[00:18:54]Ok? Now if he reacts, he will react to you like this. I am giving you a trick. The trick is if he gives you a reaction in such a way that A is reacting with 5 moles of B at 1 mole. Now these stoichiometric coefficients are helping us to see the limiting.

[00:19:13]Limiting or access reagents. So from here the product is being formed C + 3D. Now he has given that in the reaction A is 10 moles and B is also 10 moles. Now he asked, tell me what is the limiting reagent in this reaction? This question gives if he is okay? What is the limiting reagent in this reaction? Do you know what to do? Don't do anything.

[00:19:45]This is a balanced reaction to the question given the price he has paid.

[00:19:48]This is a balanced reaction. Ok?

[00:19:51]Take the price of A which he has given.

[00:19:54]How many are there? 10. Divide it by its stoichiometric coefficient. How much did it cost? 10 has arrived. Find the mole of B. How much has he given? 10. What is the stomatal coefficient of B? 5 Divide it with him. How much did it cost? Two. So now see, the value for A is 10. The value for B is two.

[00:20:16]Now whose value has increased? The big one has come for A, right kids? It has come down for B.

[00:20:21]So this means that B for which the value is less will be our limiting reagent.

[00:20:26]Remember.

[00:20:29]So if you are given a direct question like this, then you can find out the limiting reagent in this way, which one is inside the reaction.

[00:20:35]After this you should come across Percentage Access Air.

[00:20:43]Percentage Access Air So what is its formula?

[00:20:49]Percentage of exhaust is actual air minus theoretical air divided by theoretical air. Theoretical air okay? * 100 I am getting this written because in the combustion calculations, the calculations regarding combustion reactions, this percentage of excess air is used. Ok?

[00:21:22]So what percentage of access air has been used from here? Whatever is getting involved in the system is removed.

[00:21:31]You should know this one. After this, there is a percentage conversion, children, any reaction has happened like how much reactant is formed from A, how much A was converted into B, okay, so what is the percentage of conversion, so how many moles of any reactant like A, okay, how many moles have reacted, moles of A reacted divided by how many were provided?

[00:22:08]How many were the moles given to the Fed system?

[00:22:13]How many lamps were there? How many reactions were there?

[00:22:15]How many reactions were there above and below? Mole * 100 This is where your percentage of conversion comes from.

[00:22:23]Everyone knows how to extract yield. Yield Yield Percentage. What is Yield Percentage?

[00:22:29]How much actual product was made?

[00:22:33]And theoretically we calculated and saw how much should have been made, it should have been 10 grams or 8 grams. 8 grams by 10 grams * 100, from there comes the yield percentage. Ok? Then there is a selectivity term.

[00:22:54][sound of clearing throat] Selectivity. Selectivity means how much of the desired product is produced divided by how much of the undesired product is produced.

[00:23:05]Divide the one which we did not want, the desired over undesired comes from there, children, selectivity is fine, so one is density. You already know that. I got it done, mass is equal to volume. Specific gravity has to be calculated. To find the specific gravity of any substance, divide its density by the density of water.

[00:23:40]Specific gravity will come from there.

[00:23:45]After that there is a child term specific volume.

[00:23:53]Divide this density by one and you will get the specific volume.

[00:23:59]We have already discussed it in fluid mechanics.

[00:24:00]Ok?

[00:24:11]So don't forget the formula of number of moles.

[00:24:15]Mass over molar mass. So if you want to find the value of mass from this, then multiply the molar mass by the number of moles.

[00:24:28]Ok? This is very important. Along with this, there is an ideal gas equation.

[00:24:35]Ideal gas of gases. There is an ideal gas which is a real gas. The equation of an ideal gas is PV = NRT What is P here, what is pressure, V is volume, N is number of moles, R is gas constant and T is temperature, and R is gas constant, this is universal gas constant. Ok? This is the universal gas constant. It has two-three values. Its main value that will be used will be R is equal to generally it will give you the question in kilojoules only.

[00:25:20]Eight 8.314 kilojoules per mole * kelvin kilomole * kelvin this will give the value of R.

[00:25:48]Ok.

[00:26:00]After this, percentage composition by weight, percentage composition by weight, so what is this, means what is the percentage of any thing in any total, if we have any compound, look at any one atom in it or we have any molecule of it made by mixing a and b, like ab 4, what is the contribution of B in it.

[00:26:36]Ok?

[00:26:39]Percentage composition by weight has to be calculated. So what is it for?

[00:26:44]Mass of that component divided by the total mass.

[00:26:55]Multiply that by 100. So from here we will get percentage composition by weight. Ok?

[00:27:10]So, for example, let's take its percentage composition by weight, like 100 grams is 100 kilograms of solution. There is 100 kg solution. It contains 20 kg of NaCl. Ok? 20 kg There is NaCl. So what is the total mass now, baby?

[00:27:37]100 kg How much NaCl is in it?

[00:27:40]20 kg So multiply this by 100.

[00:27:43]What will be the contribution of NaCl? He will leave.

[00:27:46]How much did it cost? 20 weight percent.

[00:27:52]Ok? So this is how percentage composition by weight is calculated. In this way we can calculate the percentage.

[00:28:04]Composition by moles percentage composition. Now where there was weight, moles have to be used. Moles of component divided by total moles * 100.

[00:28:37]Similarly, children, there is one empirical formula and one molecular formula. There are two terms, you know the molecular formula, for example, take glucose, C6H12O6, this is the molecular formula, right, there is one empirical formula.

[00:28:57]This is the simplest ratio of the molecular formula. Ok? How to remove it? Don't get into this mess. It is very lengthy. Ok? Empirical formula. What is the empirical formula?

[00:29:09]Like glucose. It is made of carbon, hydrogen and oxygen, right? 6 12 6 So find its ratio. Simplest to simplest ratio. Six commons arrived. Have you arrived? 1 2 1 So here you can write six.

[00:29:26]How much carbon is there? Now the forest. How much hydrogen is there? Two. How much oxygen is there? Forest. So, this will come to the empirical formula of glucose.

[00:29:35]Empirical formula and this will come to molecular formula.

[00:29:39]Ok? This is asked a lot in exams. So, this is what you should know.

[00:29:59]clear? After this you know the composition of air.

[00:30:08]Nitrogen constitutes the most in the composition of air at 79% and oxygen at 21%.

[00:30:11]After this there is a formula, the formula for air requirement.

[00:30:20]Air requirement has to be determined. So how much air is required? Air, then whose formula will be equal to it? Oxygen Required Multiply that by 4.76.

[00:30:33]Air requirement will come from here.

[00:30:40]Ok?

[00:31:06]So I also got the formula for percentage convergence written.

[00:31:09]You are also done with the yield. Limiting access reagent also happened. Access Air also happened. Ok? That's it kids. There is nothing else. That's all. There is no need to remember anything else. Now we will ask questions about it. We practice there.

[00:31:23]Ok? Yes.

[00:31:33]Now children ask questions about this.

[00:31:36]You are sufficient enough to note this down in small pieces.

[00:31:38]Ok? You have to calculate the percentage convert yield as written, access air limiting reagent. Ok?

[00:31:47]Remove it directly from the question. limiting reagent.

[00:31:52]What is ppm? Molarity, molarity, mole fraction, number of moles, mole concept, temperature, energy, pressure and volume, area, length. Okay, this is simple unit conversion and on the basis of their stoichiometry, further conversion yield is calculated, all these things are obtained. Along with this, when you will do its questions in revision and practice those ones, then I will add some tough questions in it or the questions will be of the level which come in your exam.

[00:32:27]Ok? As much as needed. And the rest will be covered in the previous year questions that we will discuss. So there is no need to worry. Just remember as much as you have just got done and note it down in your copy. Now we will try applying them. The unit will not come.

[00:32:44]He will not ask this directly.

[00:32:45]How many grams are there in 1 kilogram? Ok? The formula of the ideal gas equation is PV = NRT, you will not ask like this. He will get you to apply. Now we will try applying it in questions. Ok? So see you in the next video lectures. We will ask questions from this chapter. That's all I wanted to tell you in this. And along with this, this is your last chapter in which only this much is required. I will take you deeply.

[00:33:09]We don't have that much time now. Ok?

[00:33:11]And anyway, there is no point in studying so deeply something which you do not need for your exam.

[00:33:16]For now let's focus on other things. So this is our 12th chapter. Here our syllabus is Finnish. After this you will get video lecture of the ninth chapter.

[00:33:30]Ok? Ninth. Yes. Questions of the eighth chapter are over.

[00:33:36]We have started the questions of the ninth chapter, right? I think we have done 20 questions of this.

[00:33:39]You got that lecture. Now we will do the remaining questions. And after this, many questions from Chemical Thermodynamics are covered in this one chapter. So let's focus on questions of chemical thermodynamics. Let's look at questions on fluid mechanics. So we just have to practice the questions, we don't have to do anything else now.

[00:33:56]Ok? So see you in the next lecture. Thank you very much.