Limiting Reagents and Percent Yield

Hinzugefügt: 2026-05-09

[00:00:16]all right hey guys at home thanks for joining us um we're kind of hit the road running today um so we are finishing up stoichiometry notes today um we're going to be looking at limiting reagents and percent yield limiting reagents is one of those honors topics that we deal with sorry i still got a couple people popping in from the lobby here we go so limiting your agents is one of those honors topics that we deal with um it's not really anything truly new calculation wise it's going to be using the same calculations we've used before um but looking at problems slightly differently so we're going to get through the notes today we'll do at least one example of each if you guys feel like we need a second one we'll do a second example and then we're going to move over and do the lab setup portion and then you'll kind of have the work time afterwards to work on your homework that's due tomorrow so remember it's not due on tuesday this week it's due on wednesday so there's only a couple problems left if you've been keeping up but that way you have a little bit extra time so looking ahead with limiting reagents so any chemical reaction as a whole if we have an insufficient quantity of any of the reactants so if we have less of one of the reactants than the other it can actually limit the amount of product that forms so if we run out of one of our reactants the reaction is going to stop running so this is our limiting reactant or limiting reagent it's the reagent that determines the amount of product that can be formed from a reaction so this will run out and be used completely up and once we run out of that reactant the reaction ends it can't make any more product than what it already has because it has to have both reactants to make products um whatever does not run out the other reactant is our excess reagent um so this is not completely used up in the reaction so some of it reacts and then you're gonna have leftovers and whenever we're finding the limiting reagent we have to convert both reactants into the same product so that they can be compared so up to this point with stoichiometry we've been given only one reactant to start with we've been given one amount with limiting reagents it will give us both reactant amounts so we have to take both of them and change them into products and we're looking for whichever one runs out so whichever one produces the smallest amount and that's how much can be made because that's going to be kind of our limiting reagent once that one runs out the reaction's done so the stoichiometry that we've been doing is the exact same but since it gives us two starting amounts we're gonna have to do two problems and we're gonna be looking for the least amount because that's how much is produced and then the reaction stops all right i think everybody's with me alright so let's look at this problem um and if you're following along on the powerpoint i did add this secondary question so i kind of added a question after this one um but once know what's the maximum number of grams of copper one sulfide that can be formed when 80 grams of copper reacts with 25 grams of sulfur and how much excess remains and i gave you the reaction of two copper plus sulfur produces the copper one sulfide and i'm just going to draw our little you're right thanks sorry so um i'm going to draw our little double road map so moles can be converted into particles grams or liters and remember we can convert that into moles of another substance which can be converted into particles grams or liters so this is two questions i added that second question on there of how much is excess because there is one like that on your homework and i want to show you how to approach um kind of a secondary question after you found the limiting agent so i got a couple people writing the problem so i'm gonna pause and let them kind of get with me box so kind of looking at what we're going to be doing so it wants to know the maximum number of grams of the copper one sulfide so we're looking for grams as our final answer and it gives us grams of copper and grams of sulfur so we're going to be changing from grams into into grams that's the path we're gonna have to take since it gave us both reactants the copper and the sulfur we're gonna do a conversion with both of these so we're gonna do them separately so we have the 80 grams of copper and the 25 grams of sulfur so look at these kind of one at a time so looking at the grams of copper first we're going to be changing from grams of copper into grams of that copper one sulfide so our first step is to go from grams to moles always so where does grams need to go in our next step yep down to the bottom so we can change into grams of copper from grams of copper into moles of copper remember grams per mole is always out of one mole and grams is the molar mass so copper's mass on the periodic table is 63.55 so that's going to allow copper to cancel with grams and leave us with moles of copper so that got us to the first m then we have to change from moles of copper into moles of the copper one sulfide so moles of copper needs to go down to the bottom and we can change into moles of the copper one sulfide so remember the mold mole step you're going to use those coefficients from your balanced reaction so that will cancel out moles and now we have moles of copper one sulfide but we're looking for grams of copper one sulfides we can add one more step on here so moles we don't need anymore so it has to go down to the bottom and we can change into grams on top so grams we need molar mass so we say copper had a mass of 63.55 but there are two of them so make sure to double that mass and add to it sulfur is 32.07 so that's a mass of 159.17 that's gonna allow moles of the copper one sulfide to cancel leaving us with grams so that's all we need to do for that problem so calculate remember anything on top we multiply by anything on the bottom we divide by so 80 divided by the 63.55 divided by 2 times the 159.17 so this will make 100 point two grams of copper or yeah i probably shouldn't circle it yet because we still have to calculate the other one so we have enough copper to make about a grams of the copper one sulfide so that's a pretty decent amount but we don't know if we have enough sulfur to make more of that copper one sulfide or less of the copper one sulfide so we also have to calculate this with the sulfur so we're going to go through the same process we don't need grams of sulfur anymore so that has to go down to the bottom to change into moles of sulfur so grams per mole is out of one mole and the molar mass of sulfur is that 32.07 so that cancels out grams now we have moles of sulfur so to get to grams of the copper one sulfide we first have to get to moles of the copper one sulfide so where should moles of sulfur go in our next step yep down to the bottom so change into moles of the copper one sulfide so looking at our coefficients there's one sulfur one copper one sulfide so that cancels out moles of sulfur and then we have moles of copper one sulfide but we need grams the nice thing is limiting reagents this last step is always going to be identical since we're changing into the same substance because moles will go down the bottom and mass will go on top so we can just copy that step down because one mole is going to be that 159.17 grams so that will leave us with grams of the copper one sulfide so then we can calculate how much can be produced based off of our sulfur so 25 divided by 32.07 times 159.17 [Music] is 124 grams of copper one sulfide so the sulfur is going to be able to make more product but we said the maximum number of product is whatever the lowest number is because once that's produced we're out of one of our reactants and the reaction stops so our answer is that 100.2 grams of copper when sulfite is what can be reduced because that's all the copper we have once we hit that 100.2 we're out of copper so the rest of the sulfur can't react it's going to be leftovers so that addresses that first question of what's the maximum number of grams of copper one sulfide that can be formed here's where that secondary question comes in how much excess remains i added this because like i said i wanted you to see what happens when you have a secondary question after you found the learning agent so we want to know how much excess remains so we want to figure out our excess is the sulfur so we want to figure out how much sulfur is left over that did not react one thing that you'll come across any time you have a secondary question whether it's how much excess is left over or how much of the other product is formed something like that you always start with your limiting reagent so we're going to start with that limiting reagent that 80 grams of copper and we want to change it into that excess reagent so we want to change it into grams of sulfur that way we can subtract from that original amount to see what is left over so we're still doing the exact same steps we're going from grams all the way to grams but we're changing from the limiting reagent into our excess reagent so we can actually use pretty much all the numbers are already up here in these upper problems so we need to first change from grams to moles so to get rid of grams of copper that goes down on the bottom to change into moles and i'm going to scroll up just a hair so my guys in the back can see that's a little too much without my guys in the back can see a little bit better so one mole is the 63.55 so we have that mass in our first reaction if i can write yeah then we need to change from moles of copper into moles of sulfur so that mole ratio step so since we have copper it's gotta go down to the bottom and change into moles of sulfur so with our reaction we have two coppers for every one sulfur and then we said we're changing into grams of sulfur so moles of sulfur needs to go down on the bottom and that molar mass was that 32.07 so that allows moles of sulfur to cancel so this is going to tell us how much sulfur reacted with the 80 grams of copper so we'll do the 80 divided by 63.55 divided by 2 times 32.07 so this is 20.2 grams of sulfur that react so to figure out how much excess we have left over we want to take our original amount that 25 grams and subtract how much reacts that 20.2 grams so there should be about 4.8 grams so 4.8 grams left over that can still react if you have more copper to react with so like i said this is an example of a problem that might come up on your homework um like i said if it was something like a double replacement reaction it might ask you how much of the other product is formed so any secondary question you're always going to start with the limiting reagent so i know that was a lot because that was two questions in one um do you guys want me to do another limiting reagent problem i'm going to take silence as a yellow good okay there are other problems if you need to go and try these like i said it's the exact same stoichiometry that we've been doing but you're solving for both reactants and you're looking for the smallest amount of product okay so let's flip over to percent yield so all right so percent yield this is something that we use a lot especially in a lab setting um so we're dealing with actual yield and theoretical yield so theoretical yield is going to be our maximum amount of product that could be formed from the given amount of reactants this is going to be our calculated value so this is our answer from a stoichiometry setup there we go so whenever you are taking the amount of reactants and calculating how much product should be formed that is your theoretical yield this also tends to be in grams is what we want to change into you're also going to deal with your actual yield so this is what you get whenever you perform a lab you make the product throughout the lab process put it on the scale to measure how much you made so this is the actual amount formed or produced in a lab experiment a lot of times this tends to be less than the theoretical so a lot of times we end up making less than we think we should be able to make so this is going to be a given value it's going to tell you like this much is produced or is found or is made in labs something like that and percent yield is going to be our ratio of our actual yield what we made in lab compared to our theoretical yield that we mathematically should be able to make and then times 100 to make a percent so here's our formula this percent yield is equal to the actual over the theoretical times 100 and this is usually going to end up being less than a hundred percent because a lot of the times not all of it reacts maybe we didn't leave it in long enough things like that um there are instances where it can be over 100 but that tends to mean that there's extra stuff in there um like maybe it's what it has water and you didn't dry out all the water or something like that but majority of the time it will be less than one hundred and sometimes you'll see this as two separate questions sometimes it'll ask you to find the theoretical yield and then a second question will ask you to find percent yield um ones that tend to be a little bit more challenging are whenever it's all within one question so we're going to do an example of one of those because those tend to be the harder ones because you're going to be given two values and you have to figure out which one is your actual yield and which one you need to use to start your calculations for theoretical yield all right so if you're following on the notes i'm gonna skip to the last slide that has an example that has both of them in the same question i think everybody's with me all right so let's look at this question um and i'll write the reaction for you guys because normally the reactions will be given plus you um so we want to know what is the percent yield if 4.65 grams of copper is produced when 1.87 grams of aluminum reacts with an excess of copper ii sulfate so i'm going to go ahead and write our reaction out okay so here's our reaction of we have two aluminum plus three copper two sulfate produces three copper and the aluminum sulfate so here it gave us both numbers and it's asking for percent yield so we need to figure out what is our actual yield and then the other number is what we're going to do stoichiometry with because in order to be able to compare two numbers they need to be in the same unit so i said for actual yield you're going to be looking for some keywords of how much was obtained or was found in lab or was created or made or produced so here it tells us that the 4.65 grams of copper is produced that is going to be our actual value so we're kind of leave that alone until we're needing to calculate that percent yield our other value the grams of aluminum is what we're going to change into our theoretical values so this is our reactant of what we're starting with and we want to look at how much product should we be able to make based off of the overall reaction and stoichiometry because in order to plug into percent yield they have to be in the same unit so we need to change grams of aluminum into grams of copper so that we can compare it to that actual yield so we're going to start with grams of aluminum so the 1.87 grams of aluminum and feel free to draw that little mole roadmap if it helps you so our first step is always to change into moles so we don't need grams of aluminum anymore so where does it go in our next step yep down at the bottom and we can change into moles of aluminum so remember grams per mole is always one mole and grams of aluminum is 26.98 is its mass on the periodic table so that's going to cancel out grams of aluminum and now we have moles of aluminum so here's where we can change from moles of one substance into moles of the new substance so since we're looking for grams of copper we need to first get into moles of copper so moles of aluminum has to go down to the bottom moles of copper can go on top so remember mole to mole use your coefficients from your balanced reactions there's two aluminum for every three copper so that's going to cancel out moles of aluminum leaving us with moles of copper and then we need to change to grams so that it can match that actual yield so theoretical yield generally always change into grams so moles of copper goes down the bottom grams goes on top so grams is always out of one mole and copper's molar mass is that 63.55 so that cancels out moles leaving us with grams so this is going to give us our theoretical yield so we're going to do the 1.87 divided by 26.98 times 3 divided by 2 times the 63.55 so this is going to be 6.61 grams of copper so we said that's our calculated value that is our theoretical yield so where we're finding percent yield which is what it's asking so we're still not done we said it's the actual so the 4.65 divided by the theoretical yield our calculated value always goes on the bottom so 6.61 times 100 to make it a percent and that's going to be 70.3 percent yield so that means that in lab we only made 70 of what we should have so this might be maybe all of our aluminum wasn't in the solution so not all of it reacted maybe it wasn't pure aluminum uh maybe we have excess water in there um maybe we didn't let it sit and run long enough so not all of it reacted there's kind of a whole variety of things that it could have caused that difference plus you know with a piece of aluminum maybe there's aluminum in the center and only the aluminum on the outside reacted there's tons of things that could have occurred to cause that percent yield um and sometimes you might get really really small numbers for percent yield that is possible um it just depends on what values you're given um i have a hand up from kendall um so because we went from grams of aluminum to the grams of copper does that mean that that was the excess that it's asking about um this is just like how much is produced so it told us that we had an excess of copper two sulfate so aluminum in this case would be like our limiting reagent um so this is how much product can be made based off of how much of the limiting reagent we have okay so we don't have to find excess in this one no only if it like specifically ask for it so that is olive percent yield so i said not a ton of new stuff today so limiting your agent it's the same stoichiometry math but you're doing it for both reactants and you're trying to figure out which one makes the least amount for percent yield you do the stoichiometry to figure out your theoretical yield and then plug it into that percent yield equation does anybody want me to do another percent yield example okay so hannah i'm going to stop recording and then we can keep going