 Hi students, today's topic is stoichiometry and limiting reagent. So before we start with the topic, we need to know what is stoichiometry. So coming to the definition, stoichiometry is the quantitative relationship. As you can see here, it is a quantitative relation between various reactants and products in terms of mass, volume, mass molecules. So we need to understand the statements in terms of a reaction first. For example, we have A plus 2 moles of B giving 3 moles of C. So how are this A, B and C related to each other? We can say that 1 mole of A is reacting with 2 moles of B to give 3 moles of C. Now as we can relate it with moles, we will be able to relate it with volume, mass and molecules as well. So this is meant by stoichiometry. It is nothing but all the reactants and the products will have a quantitative relationship between them. Now this entire topic, stoichiometry and limiting reagent, I have classified into 2 cases. First is case 1. Case 1 is, now mainly in stoichiometry and limiting reagent will be given numericals. So how to identify that when we will be applying stoichiometry concept and when we will be applying limiting reagent concept. So there are 2 cases. First case is when the information is given about only one reactant or one product. That time we will be following the stoichiometric concept. We will come to limiting reagent later. First we will be understanding how to go proceed with the stoichiometry numericals. Now we will be understanding with the help of one example. Let us say we have this particular equation. CH4 plus O2 is giving CO2 plus H2O. This is the combustion reaction of methane. Let us balance it 2H2O. Correct? So this is the equation that has been given to us. Now according to stoichiometry, I can say that, correct? According to stoichiometry, I can say that 1 mole of methane is reacting with 2 moles of O2 to give 1 mole of CO2 and 2 moles of water. So this is the basic relation between all the reactants and products in terms of moles. Now if I have related it with moles, then I can relate it with 3 other things. 1. 2. 3. 4. 5. 6. 7. 8. 9. 9. 9. 10. 11. 12. 12. 13. 14. 15. 16. 17. 18. 19. 19. 19. 19. 20. 21. 22. 22. 22. 23. 23. 24. 25. 26. 27. 29. 30. 30. 29. 50. 26. 29. 30. 9. 20. 21. 20. 21. 22.숸. 22. 23. 26. 26. 28. lifestyle. 29. So 1 mole of CH4 will be equal to how much mass of CH4 moles is 1, molecular weight of CH4 is 12 plus 4 that is 16, so it will be 16 grams. So similarly we will find for the other 3 also, 2 moles of O2 will be how much, 2 moles of O2 will be 2 into and the mass of O2 is 32, 32 grams. So 2 moles, 1 mole of O2 weighs 32 grams, so 2 moles will be weighing 64 grams. So I can say that 16 grams of methane is reacting with 64 grams of O2 to give 1 mole CO2. So 1 mole CO2 means how much mass, 44 grams of CO2 and 2 moles of water, 1 mole is 18 grams, so 2 moles will be 36 grams. So I can say this is the relation in terms of our mass okay, this is the relation in terms of or we can say is the stoichiometry in terms of mass. Now we have to find out the stoichiometry in terms of molecules, again same thing, 1 mole contains Na molecules. So I can say Na molecules of CH4 will react with 2 Na molecules of O2 to give Na molecules of CO2 and 2 Na molecules of water. This is the stoichiometry in terms of our molecules. Now what it will be in terms of volume, 1 mole is equivalent to 22.4 litres. So it will be 22.4 litres of methane reacts with 2 into 22.4 litres because this is 2 moles, so it will be 44.8 litres of O2 to give 22.4 litres of CO2 because it is 1 mole plus 44.8 litres of H2O. If you have understood this entire table, then you have understood stoichiometry. Now each one of them I can relate among each other. I can say that 16 grams of CH4 will give 44 grams of CO2, I am relating CH4 and CO2 okay. So 16 grams of CH4 will give 44 grams of CO2. I can say 16 grams of CH4 will give Na molecules of CO2 and also I can say 16 grams of CH4 will give 1 mole of CO2 or I can also say it will give 22.4 litres of CO2. Like this anything on this particular table can be connected or related to anything. I can say 64 grams of O2 will give 44.8 litres of H2O or I can say 22.4 litres of CH4 will be reacting with, this is giving, why giving? Because H2O was in the product side and O2 was in the reactant side. Now I can relate to reactant also 22.4 litres of CH4 will be reacting with 2 Na molecules of O2. So if you have understood this entire correlation, this is what stoichiometry is. Now coming to the questions, let's say they have given a question like how much volume of CO2 at STP is evolved on heating 200 grams of CaCO3. Now whenever we see, you can see here only the information is given regarding only one particular reactant or product. Any other information can you see? No. Only they have given the information regarding calcium carbonate. First things first, we have to write the formula. I mean the equation. CaCO3, heat, we are heating it. So definitely when we heat calcium carbonate we get calcium oxide plus CO2. This is our equation. So I can easily see that I will first write down what they are asking, how much volume of CO2 at STP from mass. So I will write the relation of mass and relation of volume. So mass. So here I can see one mole of CaCO3 is giving one mole of CaO and one mole of CO2. What is the mass of calcium carbonate? 100 grams. So 100 grams of calcium carbonate is giving how much of CaO? This will give 44 grams of CO2. That I know. Okay I am not concerned with calcium oxide right now because they have not asked that in the numerical. So I can see in terms of mass 100 grams will give 44 grams okay very good. So I can say 22.4 litres can give how much and also CaCO3 is not a gas. So we can ignore this part although if we had to connect then this was the correct form of the volume. Now one mole of CO2 is nothing but 22.4 litres. So I can clearly see that I can relate this with this that 100 grams of CaCO3 is giving 22.4 litres of CO2 at STP. Now simple unitary method 100 grams is giving 22.4 litres. So 200 grams will give how much? 22.4 divided by 100 into 200 that will be our 44.8 litres simple this will be our answer. So 200 grams of calcium carbonate on heating will give us 44.8 litres of CO2 at STP. I hope you have understood this. One more thing we will do let's write one equation for that let's say glucose C6H12O6 we are adding oxygen into it and it will give us 6CO2 plus 6H2 and here we will have 6O2 correct. So this is the reaction of glucose photosynthesis equation if I am not wrong yes correct only right. Now they are asking me that I want to produce 88 grams of CO2. So for producing 88 grams of CO2 how much glucose in mass will be required? How much of glucose will be required? Simple again we have just to write the relation stoichiometry in terms of mass that's all we need to do. Now first we will see that glucose what is the mass of glucose 180 grams 180 grams of glucose gives 6 into 44 grams of CO2. One mole will be how much 44 so if it is 6 moles then it will be 6 into 44 no need of calculating it right now wait 180 grams is giving 6 into 44 grams of CO2. So we can reverse it and say that 6 into 44 grams of CO2 is being produced by 180 grams of glucose so 88 grams of CO2 will be produced by 180 upon 6 into 44 into 88. Now that's why I told you don't calculate this see this is cancelling so 2, 3 so 60 grams of glucose so this 88 grams of CO2 is produced by how much grams of glucose 60 grams of glucose. How do we need to approach stoichiometry questions if you know this table whatever they are asking if they are asking volume from mass then you have to write these two columns rows. So once you write the correct relation correct correlation between each one of the reactants they are asking your job is done then you have to apply simple unitary method to proceed further. I hope it is clear so this was our case 1 where information is given about only one reactant or one product this was about stoichiometry. Now coming to limiting reagent so you can see case 2 here information regarding 2 or more reactants they have given what to do in that case that case we will be applying limiting reagent concept okay limiting reagent concept. Now what is limiting reagent so this is our case 2 so seeing the numerical if one reactant or product information is given that comes under stoichiometry we will not put limiting reagent concept there but whenever we see that information is given regarding 2 reactants 2 or more reactants then definitely there will be limiting reagent concept applied. Now what is limiting reagent that we need to know the reactant or reagent we can say which is consumed first which is consumed first which gets over first in a chemical reaction that is called a limiting reagent okay that is called a limiting reagent. For example you are eating something in India we generally eat roti chapatti sabzi. So suppose sabzi is getting over first then our sabzi will be limiting reagent but if roti is getting over then roti will be our limiting reagent simple whichever is getting over first in a chemical reaction that will be our limiting reagent. Now how to find out the limiting reagent that we need to know so suppose we have been given a plus b let's say 2a plus 3b is giving c okay like this they have given. Now let's say they have told in the question that they have given you 5 moles of a and let's say 6 moles of b they have given you. So now how will you find the limiting reagent which one is consumed first. So here you need to find moles upon stoichiometric coefficient these coefficients the coefficients that we put for balancing those are known as stoichiometric coefficients. So now n upon stoichiometric coefficient we have to find for both for a as well as b and whosoever's n by stoichiometric coefficient so let's say I am telling n upon stoichiometric coefficient of a is lesser than that of n by stoichiometric coefficient of b that will not be correct in this I think but let's see if this is true then a is the limiting reagent so whosoever n upon stoichiometric coefficient ratio is less that will be our limiting reagent. Now let's see what happens here 2a so what is the stoichiometric coefficient of a that is 2 how many moles of a is given 5 so what will be n upon sc for a it will be 5 upon 2 that is our 2.5 let's find out for b stoichiometric coefficient for b is equal to 3 and nb is given as 6 so n by sc of b will be equal to 6 upon 3 that is 2. Now clearly we can see that n upon sc of b is lesser than that of n upon sc of a. So which one will be the limiting reagent here b will be the limiting reagent so that means whenever we will be reacting these 2 b will get over first b will be consumed first so whichever is the limiting reagent that will be consumed entirely in the reaction and this a will be left in excess a will be left in excess okay so whosoever n by sc is more for this a is more that will be left after the reaction is over okay now we will see the questions regarding this let's say first one question 2so2 plus O2 is giving 2so3 this is the equation now they have told they have given you 6 moles of SO2 and they have given 32 grams of O2 O2 okay 32 grams of O2 so now they are asking which one is the limiting reagent find limiting reagent how will we do see first you need to find moles okay but this one is given in mass so find moles what will be moles moles will be equal to w upon mw so n will be equal to 32 upon molecular weight of O2 is 32 again so it is 1 okay 1 now now we have to find out n upon sc for SO2 what is the number of moles given for SO2 6 moles so 6 upon what is the stoichiometric coefficient of SO2 2 so it is equal to 3 now n upon sc for O2 we need to find out number of moles just now we have found out 1 mole 32 grams is equivalent to 1 mole so 1 upon what is the stoichiometric coefficient 1 again so it is 1 so whose whose is lesser O2 1 is lesser than 3 so this O2 will be our limiting the reagent okay so that means what you can say 1 mole we have given given 1 mole of O2 so 1 mole of O2 will react with how many moles of SO2 we can see in the balanced chemical equation 2 moles of SO2 it is reacting with 2 moles of SO2 so I can say that 1 mole is reacting with 2 moles so from 6 moles from 6 moles of SO2 2 moles have been consumed in the reaction so how much is left 6 minus 2 that is 4 moles so in the reaction mixture 4 moles of SO2 will be left because that has not been used up and we can see here 1 mole of O2 plus 2 moles of SO2 will give how much of SO3 2 moles of SO3 2 moles of SO3 so we can clearly see that in the reaction mixture in the reaction mixture in the end what will be our components that will be left whichever is limiting reagent that will be not there at all because that has been already consumed so the excess thing will left will be left reactant that has been in excess so 4 moles of SO2 will be there plus the product will definitely be there plus 2 moles of SO3 will be present so sometimes they can ask you what will be the composition of the mixture after the reaction is completed so this will be your answer this will be your answer so this was all about our limiting reagent now one more few more questions we will be doing for example let's say we can do this question suppose 2H2 plus O2 is giving 2H2 now we are been we have been given with 10 grams of H2 and 64 grams of O2 and they are asking how many moles of water will be produced in this reaction when these two will be reacted how many moles of water will be formed okay now first things first we need to find out the corresponding moles so weight upon molecular weight of H2 is 2 so this is 5 moles and for oxygen it is 64 upon 32 that will be our 2 moles now we will find out N upon SC of H2 which will be equal to 5 upon 2 that is our 2.5 and N upon SC of O2 will be 2 upon 1 this 2 moles upon stoichiometric coefficient that is 1 that will be equal to 2 now we can clearly see N upon SC of O2 is lesser so here O2 will be our limiting reagent and H2 it will be left in excess now once we find out the limiting reagent now we will go with the stoichiometry taking this as the only information so now what is given 2 moles 2 moles of O2 will be completely reacting now from the equation stoichiometry relation we have seen that 1 mole of O2 will produce 2 moles of H2 2 moles of H2 that we have seen from the stoichiometry now 2 moles of O2 will give how much moles of H2O definitely 4 moles of H2O 4 moles of H2O okay now we need to find out let's say H2 how much of H2 is left in excess so 1 mole of O2 reacts with how many moles of H2 2 moles 2 moles of H2 so now tell me 2 moles of O2 will be reacting with how many moles of H2 it will react with 4 moles of H2 so I can say out of 5 moles 5 moles was given to us initially correct 10 grams was given to us 10 grams is equivalent to 5 moles so out of 5 moles 4 moles of H2 has been consumed in the reaction so how much of H2 is left 5 minus 4 that is our 1 mole so if they are asking composition of resulting mixture then what will be present limiting reagent will not present will be not present so O2 is not present we know that H2 will be present how many moles 1 mole and water will be present how much of water was formed 4 moles of H2O so this will be the composition of the resulting mixture that is there now I hope this particular stoichiometry and limiting reagent is clear to you now there are certain questions which are important we will be discussing those the first question is read the question first try it on your own then we will be solving what is the question assuming that petrol is octane okay petrol is octane and has density so density is given as 0.8 gram per ml for what for octane volume of petrol which is undergoing complete combustion is given okay V is given as 1.425 litres so first things first can we find out that how much of octane is undergoing combustion in mass in terms of mass that will be easy so density is equal to mass upon volume so mass will be equal to density into volume that will be 0.8 into 1.425 this will come in grams okay now 1.425 again we have to multiply it with 10 to the power 3 because this is in gram per ml so if I want my answer in gram then definitely this litre I need to convert it into 1.425 into 10 to the power 3 millilitre now upon calculating this we will get the answer as around 1140 grams so one data we have deciphered that how much of petrol that is octane is reacting 1140 grams okay fine so first then I will write the equation combustion means it will react with O2 giving CO2 plus H2O now here I would like you to know one more concept here that will be very very important this is a combustion reaction and the generalized formula for combustion reaction is Cx Hy plus x plus y upon 4 O2 giving x CO2 plus y by 2 H2O now see you can easily balance it very very quickly so here it is 8 is our x 18 is our y so here it is x so here we can put 8 here it is y by 2 y by 2 means 18 upon 2 that is 9 and in O2 how much we will be putting 8 plus 18 upon 4 so this is 9 upon 2 so 8 to the 16 so 25 upon 2 so here we can put 25 upon 2 so this is how this is a very important concept it will be required later as well combustion generalized formula okay now so what they are asking on complete combustion will consume x moles of O2 okay so according to the general equation I can say that 1140 grams of what of octane is requiring how many moles of O2 requires 25 upon 2 moles of O2 okay 2 moles of O2 did I write it correct no this will not be equal to 20 25 upon 2 moles of O2 what will be equal to 25 upon 2 25 upon 2 moles of O2 the molecular weight of this C8H18 let's find out this is the thing that is given to us okay so if we find out C8H18 mass it will be around 114 grams so we can say that we can say that here 114 grams because in the equation also there is one mole here there is one mole here will be 114 and here will be one mole so we can easily say that one mole of octane is reacting with 25 upon 2 moles of O2 so 114 grams of octane is reacting with 25 upon 2 moles of O2 so 1140 grams that is given to us just now we have found out will be reacting with 25 upon 2 divided by 114 into 1140 so this is 10 this is 5 so that will be 125 moles of O2 now what is the question is the integer type question so you can see 125 moles of O2 is our X so X is our 125 so what will be the value of X upon 5 that will be 125 upon 5 that is our 25 so our answer will be 25 I hope it is clear to you next one more question we will be doing before ending this let's do this question 1 liter of CO2 is passed over hot coke okay let's write down the equation first CO2 is passed over hot coke and what it is forming 2 CO 2 CO it is forming very good one liter we have one liter okay after the reaction the volume of the reaction becomes 1.4 liters okay find the volume of CO and CO2 after the reaction okay you can see from the reaction that one liter or maybe one mole let's take it as one mole one mole of CO2 will give you two moles of CO now let's take the amount of CO2 that is reacting as X so I can say after the reaction if X moles X liters let's say have been reacted okay out of one liter this X liters have already reacted so what will be left 1 minus X liter will be left and if one mole of CO2 is forming two moles of CO then X liters will be forming how much liters of CO two X liters of CO so I can say after the reaction after the reaction how much of CO2 will be left one minus X liters and how much of CO will be formed two X liters that's all coke will not be involved okay right now the after the reaction whatever whatever will be the volume of the reaction that has been given as 1.4 so I can say one minus X liters plus two X liters is equal to 1.4 liter now we can solve for X from here it will be one plus X is equal to 1.4 so X is coming as 0.4 liters so this X was what this X was the amount of CO2 that has already reacted already reacted so now they are asking what is the volume of CO and CO2 after the reaction so volume of CO2 after the reaction was 1 minus X so it will be 1 minus 0.4 that will be our 0.6 liters and volume of CO that was formed was two X so it will be two into 0.4 that is our 0.8 liters I hope you have understood this properly stoichiometry and limiting reagent we have done n number of numericals I'll be attaching one assignment as well all the questions these questions which we couldn't do because of the time constraints so we will practice there and definitely if we get some more time we will be having a practice session number two based on this stoichiometry and limiting reagent so that's all for today so that's all for today thank you so much see you in the next class