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0:02
hey you're on the tutor hot line with Melissa Maribel how can I help you hey Melissa I need help on question 11 and number 12 one asking for the percent yield and one asking for the theoretical yield thank you okay let's go over that so this question says the combustion of 0.37 four kilograms of methane in the presence of excess oxygen produces this amount of carbon dioxide what is the percent yield so let's just start off with identifying what are we given and
0:34
what are we finding so okay what are we given well we're given two different values so I know that we have this amount for methane by the way what methane actually is it's ch4 and then it says for the presence of excess oxygen but that doesn't tell me how much oxygen we have it produces this amount of carbon dioxide so I know that well because it produces carbon dioxide carbon dioxide must be our product since it's being produced okay I know that
1:04
much now what does that really tell me though well well one I know I'm also
1:17
to talk about and go through we're going to start off with writing the balanced equation then we're gonna find the theoretical yield by doing a bit of stoichiometry and then lastly we'll find the percent yield so I'm going to start off with just writing the balanced equation now there's several hints in this question one you do have to know what methane is it's ch4 so since I know okay that's methane ch4 and Plus this is telling me either as an excess of oxygen
1:43
and it produces carbon dioxide all of these are hints the other huge hint is just the fact that it says its combustion if it's combustion well I now know the type of reaction it is and I know all the different things that are going to be part of this equation so in a combustion reaction recall that we're going to have something known as a hydrocarbon what a hydrocarbon is it's just a molecule that consists of only hydrogen's and carbons so that's our
2:10
methane in this case and what never changes is your reactant and your products so you're always going to react this with a - that's the excess o or oxygen right o - and then oxygen will not exist just by itself it's a type of diatomic molecule meaning it exists in pairs so that's why we have that o - next it's going to produce carbon dioxide and also water just because this is a typical combustion reaction so as I mentioned this never changes
2:37
this part never changes the only thing that changes is the hydrocarbon that you're initially given with so in this case I know that I'm going to start off with our methane that's provided I'm reacting that with the excess oxygen so to and will produce carbon dioxide and water so now that we have that let's just balance this out so what I like to do with balancing is just write out every single element individually on either side for our reactants and our
3:04
products and I'm going to identify how many of each element we have on either side on the left side which is the reactants and the products side which is the right type which is the product side so in this case I have a one carbon on the left side or the reactant side and then for hydrogen's I have four right because of this subscript I know that I have four next for oxygen the subscript tells me there are two oxygens on the left I'll do the same thing for the
3:31
product side so looking near this there's only one carbon so one carbon next for hydrogen there's two do to this subscript so two hydrogens and be aware of your oxygen here some people may say okay there's only two oxygen that would be wrong because don't forget about this other oxygen so we have two plus this additional oxygen so there's actually three so three oxygens okay so just a rule of thumb for balancing always leave oxygen to the very end because it tends
4:01
to kind of work itself out or get a little bit confusing if you start there so I'd say start off with anything other than oxygen so I would typically start off with carbon however carbon in this case is balanced already so now I'm going to move on to hydrogen so okay I need I need two more here right because this is four and this is two so I'm actually gonna place a two in front so it can then multiply with that hydrogen and become four so that's exactly what
4:27
I'm gonna do I'll place the two in front now when I do that it's like I'm distributing now this two into the hydrogen and the oxygen so I know well the amount of carbons didn't change because I didn't place anything in front of this carbon so I know carbon is still one but the amount of hydrogen's does change because I multiplied two times two which now gives me for our oxygen will also change because we have two times Row one since there's no subscript there Aliso - but that's
4:56
not it because I still have an additional two here so two plus two gives us four so now we have four oxygens okay so now that we have this the last thing we have to do is just balance out our oxygens so let's keep going so then now I'm going to look back here and say well what numbers do I have to place here to multiply with this two subscript ticket me four well two times two so I'll place a two here these two would then multiply and that's going to
5:22
give me four oxygens while nothing changes for carbon since I didn't place anything in front of ch4 and then nothing changes for hydrogen since once again I didn't place anything in front of ch4 so now that we have our balance equation let's go to step two which is finding the theoretical yield all right so with this type of question I know to find a theoretical yield because well one it's telling me to find the percent yield and the formula for the percent
5:46
yield is actual yield divided by the theoretical yield so I need it now the theoretical yield is found by well identifying the limiting reactant and in this question they actually tell you what the limiting reactant is already because they tell you what's going to be produced so they only give you one amount or one yeah what amount of our reactant they don't tell you the amount 402 plus they tell you that Oh two is actually your excess reactant by just
6:15
stating excess now if you need to figure out or need to go over how to find the limiting reactant I do have a video on that you can find that right here - definitely refresh your memory because that is really important with this question so continuing on with just finding the theoretical yield what I'm going to do is I'm only going to do this one since I'm looking for the theoretical yield specifically of the limiting reactant which is methane well
6:40
I'm gonna go from grams of my reactant to then grams of my product which is carbon dioxide right because this is our actual yield I need the theoretical yields of the same product which is carbon dioxide but we have an additional step just because I'm not starting with grams I'm starting with kilograms so that's actually what I have to start off with I have to start with converting kilograms methane two grams of methane by using
7:05
the metric system and then from grams of methane I have to find my way to grams of our product or grams of co2 if you don't know how to do that once again this is stray qiyamah tree well how you're going to do that is you're just going to convert to moles whatever you're not sure what to convert to typically like 99% of the time moles is going to be your best bet since all of the conversion factors equal moles okay not in this case for kilograms but
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at least when we're starting with Gramps so what I'm going to do is I'm going to convert to moles in this case so I'm going to convert to moles and then now since I'm at moles of methane I want to do a multiple ratio to then get the moles of co2 which is our product so this just allows us to change the compound that we're looking at and then finally I want to get back to grams and I'll use the molar mass in this case to get back down to co2 so grams of co2 so
7:58
this is our plant now I'm going to keep going and let's just put everything together and do the math so starting off with the kilograms that we were given initially this is the amount of kilograms I have for methane now what I'm going to do is I'm going to just convert this to gram so this is our first step what I'm going to use is the metric system so the metric system tells us that one kilogram is equal to a thousand grams so okay I know that now
8:23
all I have to do is multiply so these two would then cancel so it's show that these two cancel and then now I'm at grams of methane now that I'm at grams of methane I can then move forward to go to moles of methane how I go from grams to moles is by using the molar mass of methane so I know I'm going to use this molar mass remember how to find molar mass all we're doing is looking at the individual masses of every single element in this compound and then adding
8:52
them together so carbon is twelve point zero one and then hydrogen is one point zero one but there's four of them so I multiply that by 4 and I get this then I'll add these two values together to get this as our molar mass so what I'm going to do is just align the unit's so they can cancel so these two units are aligned across from each other so I know that those grams and grams methane would then cancel and that brings me to moles of methane so I'm
9:18
here so now that I'm here I'm now going to move on to do am Ultimo ratio to get the moles of carbon dioxide so like I said a multiple ratio is going to take place here that's what allows you to go from moles of one compound to moles of a completely different compound what I have to do is use the coefficients so I'm looking at the balanced equation and I'm looking at the coefficients involved in our balanced equation so it'll make sense once we do this so what I'm
9:45
looking at is looking at this balanced equation there's really a 1 here so there's technically a 1 right here and there's a 1 right here if there's no other number that's being shown so then I know well there's one mole of methane and I'll align the unit's again so they can cancel so those cancel and then for every 1 mole of co2 so that's on top now that I've had moles of co2 I can then convert this to grams of co2 by using the molar mass of co2 so once again
10:15
finding the molar mass I'm just adding this together so then this is a carbon at 12.01 oxygen and then I did multiply 16 times 2 just a little note so that's how I got that 32 so multiply that by 2 and then add these two values together and this is our molar mass for carbon dioxide and I'll place that here so we're multiplying by it I know that we're multiplying by it because we're aligning it once again the units across from each other so they can cancel so
10:43
that's how you know where to place everything and then from here almost done I'm now going to multiply all of this together so I'd multiply everything right across and then I divide by that 16 point zero 5 when I do that I get this value but I always have to worry about my sig figs so in this case looking at what we initially were given we were only given three significant figures so I have to round to three sig figs so I'm going to round up to this
11:10
and remember this is three sig figs because anything that has doesn't have a decimal it's a large number without a decimal those ending zeros do not matter so I only have three sig figs in this case that one the zero and three so this is my theoretical yield okay we found the theoretical yield now let's find our percent yield because we have our actual yield and we have our theoretical yield what we now have to do is plug this into our formula and even
11:38
before that I would say beware of this look at our units our units are not the same right so this is kilograms and this is grams what we wanted to do is we always want to have the same units I could either convert these grams to kilograms or I can convert these kilograms to grams either way it gives you the same result so what I'm going to do is I'm going to convert this back down to kilograms and all I'm gonna do is just using the metric system again
12:05
I'm going to divide by a thousand and that's going to give me this amount of kilograms okay I have my kilograms down now that both of these are in the proper unit the same unit and I can continue so our percent yield formula is our actual yield divided by the theoretical yield and multiply that by a hundred since it is a percentage if you forget this which please don't forget this all you have to think of is percent yield it's where it's at all right so where it's at
12:34
that's percent yield so that's just a nice way to be able to remember this just in case you forget this on your exam so all we have to do now is plug everything in so I'm going to plug in the actual yield which was given and then I'll plug in the theoretical yield which we found using stoichiometry divide these two values and multiply by a thousand and we get ninety five point four percent that is our percent yield let's move on to the next question that
12:59
asks for theoretical yield all right so this question states the combustion of 0.12 40 kilograms of propane and the presence of excess oxygen produces this amount of carbon dioxide what is the theoretical yield so it's really similar to the previous one that we just did let's once again identify what are we given and what are we finding so we are given this amount of propane and propane that's one two no propane is c3h8 and then we are given again how much product
13:29
we are producing so we are given our actual yield is it relevant do we even need our actual yield nope so this is extra information there's no need to even use it in this case because we're not asked to find the percent yield so because we're not asked to find the percent yield the actual yield isn't necessary in this question so instead all I'm using is just the amount of kilograms of our reactant or of propane and then of course what I'm
13:59
finding is our theoretical yield the steps that we're going to take to find her theoretical yield are as follows all we have to do is figure out the balanced equation and then find the theoretical yield with using stoichiometry like we did previously so let's start off with our balanced equation once again this is a combustion reaction so that means that the only thing that changes is this hydrocarbon and everything else remains the same we're adding co2 that's our
14:26
excess oxygen and then it produces that carbon dioxide and water so doing this all over again just balancing this out remember we're going to write all of the elements on either side and figure out how many of each element we have on either side and then balance them out so looking at carbon on this side we only have three due to the subscript so I have three carbons next looking at hydrogen there's eight so eight hydrogen's and oxygen there's two
14:54
alright looking at our product side carbon there's only one carbon here four let's see hydrogen there's two because of that subscript and then oxygen remember there's actually three because we have two plus this one so three oxygens always start off with I'd say anything that's other than hydrogen or oxygen always leave oxygen to the very end because it makes it a lot easier so I'm going to start off with carbon so carbon there's three carbons here and
15:23
there's one carbon there so I'm going to place a three here so I can turn this carbon into a three so when I place a three coefficient there what's actually happening is this is distributing so this 3 is being multiplied to this carbon this 3 is being multiplied to that oxygen so we now have three carbons and we have three times this two subscript we have six oxygens but it's don't forget we don't have six we have one more so beware
15:51
so three times at two six plus one more so I'd say don't just look at one side keep looking at like oh what else is this added to is there any other oxygens hidden there and there aren't so we have seven so we have seven oxygens now hydrogen nothing changed because we didn't put anything in front of hydrogen so that's going to remain the same as two so now that our carbons are balanced I'm now going to move on to hydrogen so there's eight hydrogen here and there's
16:17
two on the product side so I have to multiply a number here to two to give me eight and that's four so I know four will be place there once again I'm distributing this so distribute distribute and then so four times two gives me eight so now I changed my amount of hydrogen so that's eight and then the amount of oxygen will also change so we have four here and then plus the additional six here so 6 plus 4 is 10 so I have 10 oxygens and nothing
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changed with carbon I didn't change this number so that's gonna remain the same as 3 ok so almost there now we're finally balanced with carbon and hydrogen now our oxygen is the last thing so I have to place some sort of value here to multiply with 2 to give me 10 so 5 times 2 will give me 10 it's a place of 5 and then from here this is now 10 and everything is balanced since I didn't change anything on this carbon or anything to this hydrogen either
17:17
so this is our balanced equation continuing on our next step is to simply find the theoretical yields so this is really similar to the previous question with finding the theoretical yield it's pretty much the same exact way of doing this where I'm going to convert the kilograms to grams now after I use the metric system of course to go from kilograms to grams I'm then gonna use the molar mass of propane and convert this from grams to then moles then
17:45
continuing on with moles I now have to change the compound so I have to do am Ultimo ratio and that will bring me to moles of carbon dioxide and lastly I want to convert those to grams of carbon dioxide so I'll use the molar mass of carbon dioxide to get to grams of carbon so the theoretical yield I know is whatever the limiting reactant produces our limiting reactant in this case because it really tells us is propane since our excess reactant intint excess
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is oxygen okay is that oates here so okay I know that since that's given I don't have to do this twice and let's keep going with this so I'm just going to put this all together so starting off with what we're initially given the kilograms of propane and I want to convert this to grams so once again I'm going to convert this to grams and I'm just using the metric system which was that one kilogram is equal to a thousand grams aligning our units across from
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each other so they can cancel so this cancels this cancels we're now at grams of propane so I'm here next I want to move on to moles of propane how do I go from grams to moles I use the molar mass so I need to find the molar mass of propane which is what I did here multiplying that 12.01 times the three subscript to give us this and then of course multiplying one point zero one times that eight subscript to give us this add these two values together and
19:08
forty four point eleven is our molar mass of propane so of course once again I want to align the unit so they can cancel so these two grams across from each other so they can cancel out and I'm now at moles of propane so I'm here at this step now I want to change the compound so when I want to change the compound I'm gonna use a multiple ratio which is found using our balanced equation okay I want to align my units so they can cancel and remember if we
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don't have a coefficient in front then assume that it's just one so in this case I know that for one mole of propane which I'll align the unit so they can cancel there's three moles of carbon dioxide which is on top so our units of moles of propane will cancel and then next I'm here so I just have to convert this to grams of carbon dioxide by using the molar mass so remember molar mass allows us to go from grams to moles or vice versa moles to grams so then
20:03
identifying the molar mass of carbon dioxide we did that previously and all I have to do instead align the units again so they can cancel moles of carbon dioxide are on the bottom those would then cancel out to get grams of carbon dioxide and we're just going to multiply everything on top so all of these out and then divide it by that forty four point eleven and it's going to give us this value we can stop here that would still be correct but because
20:29
they gave us the amount of kilograms or the amount of kilograms in general right everything was initially in kilograms they most likely want the answer to remain in kilograms so I'm going to convert this back down to kilograms and all I have to do is using that metric system again write this conversion factor is line those units across so they can cancel so I'm dividing by a thousand and then that's going to give me this amount of kilograms overall so
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this is my theoretical yield this question is a common test question and a final exam question yeah I know so make sure to study these types of questions anything talking about percent yield theoretical yield limiting reactants all of that is the most common final exam question now if you have other questions and you want the answers to them and want to figure out what am i doing please ask me simple as that so in the description books you'll find a link
21:24
and you could submit your question and I'll be happy to answer it and I'll see
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