hey all welcome to homeschool and we are with class 11 chemistry series in my previous video we have discussed everything about stoichiometry and how to calculate stoichiometric problems we have seen four different problems on four varieties isn't it and today I am with the most important topic for competitive point of view that is limiting reagent. So we will try to understand what do you mean by limiting reagent and then we will see a problem like how to identify which is a limiting reagent in a reaction. Fine and you know what limiting reagent related questions will definitely I mean I don't say definitely most of the times they have appeared in previous NEET or JEE papers. so definitely this is one of the important concepts of class 11 and 12 chemistry where i'll be teaching you a super trick to solve the problems of limiting reagent easily so watch the video till the end definitely you will have a maximum benefit from the video and after the video just share your opinion whether you like the trick or not fine so first let us understand What do you mean by limiting reagent? See its definition is very simple.
Limiting reagent is that reagent which limits the reaction. Or I can say the reagent which can stop the reaction. Limit in the sense the reagent that helps in stopping the reaction.
So, which stops the reaction. This way also you can talk about. So, I think you must have got some idea about limiting reagent, right? So, the reagent which can stop the reaction. Any reaction, say for that matter, A plus B reacting to give products.
So, here these are the two reactants. So, among the two reactants, the reactant. patent which is responsible for stopping the reaction is called limiting reagent see guys this concept I'll make you understand with a better real life example try to understand the example first after that automatically you will understand about limiting reagent see I have to make five chapatis see I want to make five chapatis for lunch or dinner so to make five chapatis my recipe the recipe that i have to follow is uh two bowls two bowls of flour right two big two big bowls of flour right and i want one glass of water one glass of water So when I take two bowls of flour, I add one glass of water. I nicely make a dough and then out of this, out of this, I can make five chapatis. Right.
So if I take two big bowls of flour and when I take one big glass of water, then I can make five chapatis out of this. right but but imagine i have only one bowl of flour and i have one glass of water can i make five chapatis my target is to make five chapatis to make five chapatis these are the conditions necessary these are the requirements right but i have only one bowl of flour And one glass of water. Water is okay.
Enough amount of water is there to make five chapatis. But you see the flour, I have only one bowl. So among flour and water, can you tell me which is stopping me to make five chapatis? can you tell me out of this maximum how many chapatis i can make maybe two or three chapatis i can make but not five see my number of chapatis is not sufficient i want five chapatis but i'm getting only two or three which is responsible for not having five chapatis for not making five chapatis the among the two the responsible guy is the flower right so flower is stopping me to make five chapatis because I have limited amount of flour here.
I have enough amount of water. I want one glass of water. I have one glass of water to make five chapatis but the flour is not enough. Flour is only one cup I have, one bowl I have. So flour is the one which is stopping me to make five chapatis.
So I would call here, the flour is called as limiting reagent. The flour is called as limiting reagent. You can make some chapatis but not five. Okay, but not...
the amount that i want i'm making less chapatis it's because the flour is less flour is stopping me to make five chapatis okay say that flour which is limiting uh to make five chapatis which is stopping me to make five chapatis is called with the name limiting reaction and this guy is called accessory agent Excess. I have nice excess of water. More amount of water I have but very little amount of flour I have. Flour is responsible to stop the reaction here.
Right? So here the one which is very less than expected. Okay.
The one which is actually stopping the reaction is called as limiting reagent. Okay, so now let us come back to a chemical reaction. I told A reactant combines with B to get product.
If you want to get nice product, whatever you are expecting, both A and B should be there in a sufficient amount. If one is less, then that reagent stops the reaction, right? So because of that reagent, which is less in amount, you will get very little product. That finally reaction gets stopped.
So that reagent we will call it as limiting reagent. And the other reagent is called as excess reagent. Clear?
So hope you have understood the concept of limiting reagent. If you have understood the concept of limiting reagent, just post a comment in the comment section. Bye. So, okay, understood what do you mean by limiting the agent. Now the question is, they would ask certain problems.
Say, they'll give some reaction. some information would be given say this much amount of A reacted with this much amount of B you are getting some product and they will ask you to identify which is a limiting reagent. So some information they will give about the reactants and they will ask you which one of them is limiting reagent or which one of them is excess reagent. So we will see few questions on that. See guys, here is a very simple question for you.
5 grams of nitrogen and 10 grams of hydrogen are mixed. When you mix, you will get ammonia. Identify the limiting reagent and also calculate the amount of ammonia formed. So this is the question. So first thing is you need to identify the limiting reagent.
I will give a beautiful trick to identify. So the first thing you need to do is write a chemical equation. They said nitrogen reacts with hydrogen to give ammonia.
This is a chemical equation. Balance it. It must be balanced.
Okay. So how do you balance this? Two we will put here and this is three. Right. So now this is a complete balanced chemical equation.
And now. Okay, write the information given. So how much of nitrogen is it? Yes, 5 grams of nitrogen they have taken.
And how much of hydrogen has been taken? It is 10 grams of hydrogen. So you mixed 5 grams of nitrogen with 10 grams of hydrogen.
And they are asking which is a limiting reagent. Maybe some reagent is less in amount than expected. Some reagent may be in excess. So they are asking you to find that only which reagent is limiting reagent here.
And again, you have to find how much amount of ammonia has formed when this when 5 grams of this combined with 10 grams of this how much ammonia is formed. See, you can find the second question only if you know the answer for first question. Okay, say when they asked you how much amount of product is formed, you must know which is a limiting reagent here. If you don't know the limiting reagent, you can't find how much ammonia has formed.
So now the first step is finding limiting reagent. Okay, so I will explain how to find limiting reagent. Finding limiting reagent, LR.
We write it as LR in shortcut. That means limiting reagent. Okay, and you know what? First thing that you need to do for finding limiting reagent is find, find the value of, value of, okay, number of moles, number of moles divided by stoichiometric coefficient, stoichiometric coefficient, coefficient for both reactants, okay. For both reactants.
Clear? So you need to find the value of this. Number of moles divided by stoichiometric coefficient.
This is the formula you have to apply for both reactants. Okay, fine. So, now let us apply the formula for this and apply the formula for this also. What is the formula? Number of moles divided by stoichiometric coefficient.
What do you mean by stoichiometric coefficient? It is the digits that you have put for balancing. If I ask you what is stoichiometric coefficient of N2, then the answer is 1. If nothing is there, here it is understood that there is 1. If I ask you what is the stoichiometric coefficient of H2, then the answer is 3. Okay, fine. And you see here number of moles. Did they give number of moles directly in a question?
No, mass is given, right? They have given mass. So, this mass, you should first convert it into moles. Okay, so how do you convert that into moles? You see number of moles of N2.
N2. So how much do you get? What is the formula?
It is mass divided by molar mass or molecular mass. So what is the mass given? Mass given is 5 divided by N2's molecular mass. N atomic mass is 14. 14 into 2, 28. Right.
So what is the answer you will get? 5 divided by 28. I think you will get around 1.77. Okay. So number of moles of nitrogen is 1.77. Likewise, calculate number of moles of hydrogen also.
Number of moles of H2. Same formula you can use. The same formula.
What is the formula? Mass by molar mass. So what is the mass given? 10. right divided by molar mass of h2 hydrogen's atomic mass is 1 1 into 2 2 right so how much do you get here 1's are 5's are so number of moles you will get it as 5 so this is the number of moles of hydrogen this is the number of moles of nitrogen right now apply this formula Okay, so number of moles you want.
We didn't have in a question. They gave mass. So this mass we converted into moles using this basic formula.
That's what we have done so far. Now you apply this formula for both reactants. Okay, so what is the formula you will apply for both reactants? Number of moles.
Let us do it for nitrogen. Okay, for nitrogen. So, what is the number of moles of nitrogen?
It is 1.77 divided by stoichiometric coefficient. What is the stoichiometric coefficient for nitrogen? It is 1. Okay.
Now, let us do it for hydrogen also. Okay. For hydrogen, what is the number of moles? 5 divided by stoichiometric coefficient is the formula.
What is the stoichiometric coefficient for H2? It is 3. So, 5 divided by 3 okay so what is the answer you will get here 1.77 you will get and here 5 divided by 3 you will get around 1.6 isn't it and now select the least value select select the least value least value okay and the one which has got least value is the limiting reagent you The one with least value is the limiting reagent. Okay. So which is the least value? 1.77, 1.6.
Definitely 1.6 is least value. So which is a limiting reagent? 1.66 you got for H2.
So among N2 and H2, the limiting reagent is H2. okay so H2, H2 is your limiting reagent, got the answer for first question. They asked identify the limiting reagent. So this is what they will ask in the competitive exam. So they will not give you moles directly.
Okay. They will give you mass only. Convert that into moles using a simple formula.
Then you apply which formula? Number of moles by stoichiometry coefficient for both reactants. Okay. Calculate this for both reactants. So calculate it.
Number of moles of nitrogen we got this much divided by stoichiometric coefficient is 1. Number of moles for hydrogen you got this much divided by stoichiometric coefficient for hydrogen is 3. Okay, you will get some value, right? Among the two values calculate, I mean select the least value, less value. And the one which has got the less value is considered as limiting reagent. So this is the trick, super trick. to identify the limiting reagent okay clear fine so once you got to know which is a limiting reagent then next part of the question is much easy you should calculate the amount of ammonia formed with respect to limiting reagent okay so you should calculate the amount of ammonia formed with respect to limiting reagent Okay.
Say here from the equation. Okay. Now I'm calculating the second part of the question.
I'm calculating second part of the question that is calculating, calculating amount, amount of ammonia formed. Okay. With respect to calculate this, calculate This with respect to limiting reagent. Here the limiting reagent is hydrogen. okay so this is very important and see how i am calculating okay let me erase this one okay see we have a chemical reaction n2 plus h2 giving you 2nh3 right n2 plus 3h2 giving you 2nh2 okay now you just pick it out from the question that uh three moles of hydrogen is giving two moles of ammonia right?
So, 3 moles of hydrogen is giving 2 moles of ammonia. So, from the equation, I can write from equation, from equation, what I can write? 2, 2 moles of hydrogen is giving, sorry, 3 moles, right? Yes, 3 moles of hydrogen is giving 2 moles of ammonia. 2 moles of ammonia okay so now how many moles of hydrogen you have with you from the information given in a question okay say in the question you had only 10 grams 10 grams means how many moles we calculated moles here 5 moles right so you have 5 moles you have 5 moles so how many moles of ammonia you will get from the equation it is understood that 3 moles of hydrogen will give 2 moles of ammonia.
This is standard. Okay. In the previous class, we did stoichiometric calculations.
No, similar way. This is standard we got from equation. Now, how many moles of hydrogen is there with you from 10 grams?
From 10 grams, how many moles? 5 moles, right? So, 5 moles will give you how much?
So, just what you have to do cross multiplication. 5 into 2 divided by 3, you have to do. Okay, so just calculate 5 2's are 10 divided by 3. So 1's are 3's are right.
So 3.3 something you will get. Right. So how many moles of ammonia? You will have 3.3 moles of ammonia.
Okay, so this much amount of ammonia is solved. Okay. So 3.3 moles means say if they ask you in mass, say you are writing your answer in moles.
But if they ask you in mass, how do you calculate? You know the formula, right? Number of moles is equal to mass by molar mass, right? So number of moles you know 3.3, right?
is equal to mass if you want to calculate mass by molar mass of ammonia is how much? It is 14. Atomic mass of nitrogen is 14 and here atomic mass is 1. 1 into 3, 3. 14 plus 3, 17. Right? So if you do 3.3 into 17, you will get your answer in mass.
So this much will be the mass of ammonia. Right? So you can, according to the question, If they ask your answer in moles, then keep it. But if they ask your answer in mass, the same thing you convert it in grams that is in mass. Just if you do 3.3 into 17, you will get your answer in grams that is mass in terms of mass.
Okay. So you can calculate your answer in terms of moles or you can calculate your answer in terms of mass also. Okay.
So this is the basic problem that you can do on limiting reagent. okay so when they ask you how much amount of product is formed when this match reactant reacted with another reactant you know major thing you need to know is limiting reagent so with respect to the limiting reagent only you can calculate how much product you had got this way okay so this is the standard we took from the equation now we have five moles from the information given in a question so from the equation if this moles giving this much 5 moles will give how much so we got our answer in terms of moles but if you want to present your answer in terms of mass convert we have a simple formula right using this we can always convert moles to mass mass to moles vice versa right so that's all about limiting reagent okay so we will go for one more question look at this question guys actually this is the question I picked from one of your previous neat paper say they actually ask you uh which is a limiting reagent among mg and the oxygen okay so that was the question but i added two more questions to this i am also asking you which is excess reagent and also asking you how much mg is formed okay fine so uh let us read the question one gram of mg is burnt in closed vessel containing 0.5 grams of o2 okay So first let us try to write the chemical equation. Mg is reacting with O2 to give MgO. This is the chemical equation. And balance the equation.
Yes. Say we put 2 here and we put 2 here. Right. So 2Mg plus O2 reacting to give 2MgO is balanced chemical equation.
And let us write what information is given. Information, see they said 1 gram of Mg is reacting with 0.5 grams of oxygen, right? And to find the limiting reagent, what you have to apply the formula? You should apply the formula that, let me write the formula here, number of moles divided by stoichiometric coefficient, right?
This is what the trick formula that you have to remember. So this formula you should apply for both reactants. Whichever the case you get the least number that is selected as limiting reagent. That is your trick. Okay.
So now to apply this formula we want moles. You should calculate moles first. Okay. So how do you calculate number of moles of Mg? How do you calculate number of moles?
What is the formula? Mass divided by atomic mass or molar mass. okay so molar mass or atomic mass we know this so apply this formula and first calculate moles okay so what is the mass given 1 divided by what is the atomic mass of mg it is the element so atomic mass atomic mass is 24 so 1 divided by 24 how much do you get you will get around 0.04 yes so the number of moles of Mg is 0.04 moles. Okay. Similarly, calculate here the moles 0.5 divided by what is the molar mass that is molecular mass of O2. Okay.
Oxygen's atomic mass is 16. 16 into it will be 32. Right. So 0.5 divided by 32. So you would be getting around 0.01 moles, right? So we have 0.04 moles of Mg and 0.01 moles of oxygen, isn't it? So now we got number of moles, then we can apply this particular formula, isn't it?
So applying that formula for Mg. Okay, so what is the formula? Number of moles by stoichiometry coefficient. What is the number of moles of Mg? Just now we got 0.04 divided by stoichiometry coefficient is 2. So, we will get 0.02, right?
So, this is the number we got for Mg. Similarly, calculate for H2. Apply the same formula, number of moles by stoichiometric coefficient for O2. Number of moles of O2, how much you got? 0.01.
Divided by stoichiometric coefficient of O2 is 1. So, what is the number you will get? 0.01. So, which is the least number? 0.01 is the least number. So, definitely oxygen, oxygen is limiting reagent.
So, you got your answer, oxygen is limiting reagent. So, you have to mark for option where oxygen is present. Okay.
So, first question answer we got, oxygen is limiting reagent. When oxygen is limiting reagent, which will be the excess reagent? Definitely Mg.
The other reagent is excess reagent. Right. So, when oxygen is Lr, then Mg is excess reagent.
Mg is excess reagent. Got to know? So we got the answer for first question as well as the second question.
Now let us go for working with answer for C question. So what is the question? How much MgO is formed? So always how much amount of product is formed is calculated with respect to a limiting reagent.
Isn't it? Right? So which is a limiting reagent? O2.
Now take the information from the equation first some standard information you have to take so from the equation we can clearly say that uh you know one mole of o2 is giving two moles two moles of mgo Right. And now how many moles of O2 you have in your reaction? How many moles we got here?
0.01 moles. Right. So our limiting reagent moles which is there in a reaction is 0.01 moles. So 0.01 moles of O2 would give you how many moles of MgO just across multiplication. That is 0.01 into 2 you have to do, right?
Divided by 1. So, you would be getting 0.02 moles of MgO. Okay? So, this is your answer.
If they ask your answer in moles, it is 0.02 moles. Okay? But if they ask your answer in mass, in terms of mass, if you have to present in Your answer, then how do you calculate?
We know the formula. Number of moles is equal to mass by molar mass. right mass by molecular mass number of moles we know 0.02 into molar mass of MgO you have to calculate Mg is 24 plus 16 right so 24 plus 16 is how much it will be 40 so 40 is the molar mass into 40 if you do you will get your answer in mass mass of MgO you will get right so if you do 0.02 into 40 you will get your answer in terms of mass okay so this is one more problem i have shown you so this is how the questions can be asked and for the competitive examinations they would ask you identify the limiting reagent so just follow the trick that i taught you and you would get answer within few seconds only you no need to go for a longest way okay fine so that's all about the limiting reagent and only one last concept is remaining in a chapter that is concentration in solutions okay or you know we will study reactions in solution that's how the heading is there in a textbook but actually we will study many concentration terms all of them we will discuss in the next video and with next video i am going to complete the entire chapter clear so that's all for today and we will meet up in the next video till then take care revise and do subscribe our channel to learn the concepts in the easiest way all the best and good luck