[Music] hey all welcome to home school and we are with class 11 chemistry series uh 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 je 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 a limiting reagent is that reagent reagent which limits the reaction which limits the reaction or i can say the reagent which can stop the reaction limit in the sense uh the reagent that helps in stopping the reaction okay 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 ah which can stop the reaction any reaction say for that matter a plus b reacting to game products so here these are the two reactants so among the two reactants the reactant which is responsible for stopping the reaction is called limiting reagent so you 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 the agent see i have to make five chapatis say 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 bounds of lord i 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 am 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 reaching 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 i'm making less chapatis it's because the floor is less flory is stopping me to make five chapatis okay say that flower which is limiting uh to make five chapatis which is stopping me to make five chapatis is called with the name limiting creation and this guy is called excess reagent 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 see now let us come back to the chemical reaction i told a reactant combines with b to get product if you want to get a 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 clean so hope you have understood the concept of limiting reagent if you have understood the concept of limiting reagent just post a comment in a comment section fine 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 uh a reacted with this much amount of b you are getting some product and they'll ask you to identify which is a limiting reagent okay so some information they'll give about the reactants and they'll ask you which one of them is limiting reagent or which one of them is excess reagent okay so we will see few questions on that so you 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 okay identify the limiting reagent and also calculate the amount of ammonia formed so this is a 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 2 we will put here and this is 3 right so now this is a complete balanced chemical equation and now okay write the information given so how much of nitrogen is there yes 5 grams of nitrogen they have taken and how much of hydrogen has been taken it is 10 grams of hydrogen so you mix to 5 grams of nitrogen with 10 grams of hydrogen and they are asking which is a limiting reagent may be some reagent is less in amount than expected and 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 ask 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 startimetric 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 one if nothing is there here it is understood that there is one if i ask you what is the stoichiometric coefficient of h2 then the answer is three 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 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 is molecular mass in n atomic mass is 14 14 into 2 28 right so uh 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 one one into two two right so how much do you get here one's a fisa so number of moles you will get it as five 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 one okay now let us do it for hydrogen also okay for hydrogen what is the number of moles five 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 the one with least least value is the limiting ph okay so which is the least value 1.771.6 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 stoichiometric coefficient for both reactants okay calculate this for both reactants so calculated number of moles of nitrogen we got this much divided by stoichiometric coefficient is one number of moles per hydrogen you got this much divided by stoichiometric coefficient for hydrogen history 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 easier 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 sep second part of a question that is uh calculating calculating amount amount of ammonia formed 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 arrange this one okay see we have a chemical reaction n2 plus h2 giving you 2 nh3 right n2 plus 3h2 giving you 2nh2 okay now you just pick it out from the question that uh three poles 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 uh sorry three moles right yes three moles of hydrogen is giving two moles of ammonia two 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 bowls we calculated moles here five moles right so you have five moles you have five moles so how many moles of ammonia you will get from the equation it is understood that three 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 phi twos are ten divided by three so ones are uh threes are right so three point three something you will get right so how many moles of ammonia you will have three point three moles of ammonia okay so this much amount of ammonia is formed here so 3.3 moles means say if they ask you in mass say you are writing your answer in poles 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 one into three three fourteen plus three seventeen right so if you do three point three into seventeen 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 as such but if they ask your answer in mass the same thing you convert it in uh 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 uh 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 much 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 a standard we took from the equation now we have 5 moles from the information given in a question so from the equation if this moles is 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 need paper say they actually asked you uh which is a limiting reagent among mg and the oxygen okay so that was the question but i added the two more questions to this i am also asking you which is excessive and also asking you how much mg boy 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 2 mg plus o2 reacting to give 2 mg oh 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 uh what you have to apply the formula you should apply the formula uh 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 bolts first okay so how do you calculate number of moles of mg how do you calculate number of modes 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 one divided by what is uh 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 ah calculate here the moles 0.5 divided by uh what is the molar mass that is molecular mass of o2 okay oxygen's atomic mass is 16 16 into 2 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 stoichiometric coefficient what is the number of moles of mg just now we got 0.04 divided by stoichiometric coefficient is 2 so we will get 0.02 right so this is the number we got for mg similarly calculate uh 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 one so what is a 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 the energy 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 ch 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 uh how much amount of product has 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 mg 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 e 0.01 volts so 0.01 moles of o2 would give you how many moles of mg just across multiplication that is 0.01 into 2 you have to do right divided by 1 so you would be getting 0.0 2 moles of 0.02 moles of mg 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 uh 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 mg 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 thought you and you would get answer within few seconds only you know 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 uh 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 foreign