Introduction to Stoichiometry, Percentage Yield, Actual Yield, Limiting and Excess Reagent.

Added: 2026-05-15

[00:00:00]Hi everyone and welcome back to an interesting chemistry class. Today we are going to be exploring in full the concept of stoic geometry and in the course of explaining this concept we're going to be solving this question right before you. The question reads, "Hydrogen sulfide given off by decaying matter is converted into sulfur dioxide in the atmosphere by the reaction.

[00:00:29]2 H2S + 3 O2 react together to produce 2 S2 plus 2 H2O. Now the first question says how many moles of H 2S are required to form 8.20 moles of sulfur dioxide. Now if you want to solve this type of question there are a couple of things you need to know how to do. The first thing which is very important is that you need to know how to determine the molar mass.

[00:01:08]I'm going to say number one, please ensure you write it down. The molar mass of a compound. That's the first thing.

[00:01:20]The second thing you need to know is you need to convert you need to learn on you need to know how to convert from mass to moles.

[00:01:33]The third one is you need to know how to do what we call a mole to mole ratio.

[00:01:45]Okay. So these are the three most important skills you need to know whenever you're dealing with stoometric questions. Okay. Now, first thing that we need to ensure here is that this chemical reaction or equation is balanced. Make sure you always balance your chemical equation. If it is not, then balance it. So, let's check here.

[00:02:08]We have four cuz 2 * 2 that's a four, right? Four atoms of hydrogen on the left and four atoms of hydrogen on the right. So, for the hydrogen atoms, they're balanced. Now for the sulfur we have 2 * 1 atom of sulfur and that gives us two atoms of sulfur on the left and we have two atoms of sulfur on the right. So for the sulfur atoms they are balanced. We have six atoms of oxygen on the left three * that's a six and we have 2 * 2 that's a four.

[00:02:49]Okay. So here we have four and for this one here we have 2 * 1 atom of oxygen that's a two. So 4 + 2 that's a six. So basically we have six on the left and six on the right. So the chemical reaction is perfectly balanced. [snorts] Now let's now go to the next question which is the first one here. It says how many moles of hydrogen sulfide are required to form 8.20 20 moles of SO2.

[00:03:22]Okay. So, how do we do this?

[00:03:26]This is hydrogen sulfide and we're given the moles of SO2 that were produced already. Okay. So we know that this number here in front which is a two means that for every two moles of hydrogen sulfide that reacts in this reaction two moles of sulfur dioxide are produced.

[00:03:59]This is sulfur dioxide. So for every 3 moles of oxygen that reacts two moles of this are also produced. That is what it means. So this numbers in front are called stochometric coefficients.

[00:04:15]So they actually tell you the amount that are reacted in the amount that are produced. So right here 2 moles of H2O of H2S rather react with 3 moles of O2.

[00:04:33]Okay, they both react to produce two of this. So that is what it means. Two moles of H2S react with 3 moles of O2 to produce two moles of SO2 and H2O. Okay. So that is what it means and we use these numbers in front to do our mole to mole ratio.

[00:04:54]These numbers in front are called stoeometric.

[00:05:01]Stoometric coefficients. Okay.

[00:05:06]Stoometric coefficient. That's what they actually called. So we use them in our mo to more ratio calculation calculations. All right. So now let's solve the first one. So let's solve this question and see how we can come to a correct solution. For the first one here we have 8.2 moles 8.20 moles of SO2. Right? Okay. So the question right here is what possible amount okay of H2S can produce exactly 8.20 moles of SO2.

[00:05:55]That's the question. So we actually we have our product. We're doing a backward calculation.

[00:06:02]Okay. Since you have the product, we can do a backward calculation to determine the exact amount of hydrogen sulfide that can produce exactly 8.20 moles of SO2. Now, we're going to use what we call the chain link method. So, the chain link method will help us to find what we're looking for. It's also the perfect method uh for our mole to mo ratio. So the mole to mole ratio is more like a bridge that connects you from mole of reactant A to mole of either reactant B or our product A or product B. So it's more like a ladder that connects you between two reacting species. So right here we have the one we are looking for is H2S right? So it has to be on top. So we say in front here we have a two right.

[00:07:04]This represents two moles of H2S. We say for every two moles of H2S that reacts exactly two moles of SO2 are produced.

[00:07:22]So two moles of SO2 are produced. Now the question is why is SO2 down below? It is because we want to cancel it out and only have uh H2S remaining. So the secret there is the one you are looking for has to be on top. So let's now divide right here. 2 / 2 that will give you a one right? So we now have 8.20 20 moles or of H2S. Okay, cuz that's what is remaining. We've cancelled the SO2, right? Correct. So, times 1 equals. So, this is 8.20 mole of H2S.

[00:08:14]So, this actually makes sense. It's actually telling us here that 8.20 20 moles of H2S can produce exactly 8.20 moles of SO2 and the reason is because of the molar ratio or the mole to mole ratio because see here for every two moles okay for every two moles of H2S that reacts exactly two moles of SO2 will be formed.

[00:08:45]So if we for example we triple or we double this mole let's assume we have a four here. So that's going to be for every four moles of H2S that react exactly four moles again of SO2 will be formed. Even if it's 10 moles that react exactly 10 moles will be formed. So that is what we mean by mole to mole ratio.

[00:09:09]Now we've answered the first one. Let's go over to the second one. Here it says how many grams of oxygen okay are required to react with 1.00 mole of H2S. So how many g of oxygen are required to react. Now we also do our mole to mole ratio cuz now we want to go from the mole of H2S to the mole of oxygen. So remember the more to mo ratio is like a bridge that connects two things together or two species together in a chemical reaction. You cannot just go direct. Okay, you need to go through the bridge and that is the essence of our mole to mole ratio. So here the second one here it says uh here this is the second question. So we have 1.00 0 0 mole okay of H2S now which one are we looking for we are looking for O2 right correct so O2 has to be on top I'm going to say times you set it up in a way that the one you're looking for has to be on top now what is the mole to mole ratio one more time we want to go from the mole of H2S to the mole of oxygen you can For every two moles of H2S that react, three moles of oxygen also react. So we don't use this subscript. Don't use the numbers down there. They are called subscripts. Use the ones in front whenever you want to do your mole to mole ratio. So we say for every two moles of H2S that react. Okay. So we here we have two mole of H2S because it has to be below because we're looking for oxygen right so we say for every two moles of H2S that reacts exactly three mole of O2 reacts with it at the same time okay such that this mole here cancels with the one down here and then we now have 1 multiplied by 3 mole of O2 all over 2 right correct so this is equal to 3 /2 is 1.5 so we now have 1 0 * 1.5 mole of O2 so this is should be 150 this equal to 150 mole of O2. So this means that whenever exactly 1.00 mole of H2S reacts, exactly 1.50 mole of O2 will react with it instantly. So that is actually what it means or 150 moles of O2 is required to react with 1.00 mole of H2S. So that's the second part. Now part C. Let's go on to part C.

[00:12:24]So for part C here it says how many grams of O2 are required okay or how many grams of water are required to pro are produced rather from 6.82 g of H2S how many g of water are produced from 6.82 82 g of H2S. Okay, so this is where knowing how to determine the molar mass and how to convert from grams to moles comes. Okay, in play here we're given 6.82 82 g of H2S and we were asked to calculate how many g of water will be produced from 6.82 g.

[00:13:17]Remember that water is also a byproduct of this reaction. So we need to calculate exactly the amount of water that we that we produce from the reaction of 6.82 g. So the first thing that we need to do here is that we need to understand that you cannot jump from mass okay of one reactant to the mass of another reactant or from the mass of one reactant to the mass of a product. You can't these are the steps. First you go from g to moles or from mass to moles because we are using moles for our mole to mole ratio. We don't use grams for mole to mole ratio. Take note of that.

[00:14:01]We don't use mass for mole to mole ratio. We use moles to do that. So if you're given the mass in grams, then it is paramount or pertinent you convert from mass to moles or from grams to moles before you can start working on anything. I hope you got that. Thank you. So let's see. Right now first and foremost here we're given the grams.

[00:14:28]How do we determine the moles of H2S? So we are going to say moles is equal to mass over mass.

[00:14:40]This is a standard formula for converting from mass to moles. And then for the mar mass, what is the mar mass of hydrogen sulfide? Well, if you go to the periodic table, you will observe here we have molar mass.

[00:14:56]Molar mass. Okay. So on our table, so we go to the periodic table.

[00:15:06]This is very important. You need to know how to use it.

[00:15:10]So on that table, we know that here we have H2S, right?

[00:15:17]So H2 and then S.

[00:15:21]H is hydrogen right? So the atomic mass unit of hydrogen is 1.08.

[00:15:27]And since we have two atoms of hydrogen, you multiply it by 2. For sulfur, the atomic mass unit instead is 2.06.

[00:15:35]Okay, we only have one atom of sulfur.

[00:15:37]So you don't have to multiply anything.

[00:15:39]Now when you add them together, we are going to have 1.8 8 * 2 + 32.06.

[00:15:49]And when you add them together, you're going to get 34.08 g per mole. So this is the mar mass of hydrogen sulfide. So we now divide the mass by the mar mass to get the moles.

[00:16:03]We say moles is equal to what's the mass? The mass we're given is 6.82 82 uh g of H2S and then over the mar mass is 34.08 g per mole. So when you divide these two what you are going to get is 0.20.

[00:16:26]So the moles I'm going to say n. So we use n to represent moles. Okay. So moles of H2S equals remember that this N represents moles. Okay. of H2S is equal to 0.20 2 0 mole. Okay. So this is exactly the uh the amount of substance contained in 6.2 6.82 g of hydrogen sulfide. Now having converted from g to moles we can now use this moles to work our way through. So what we do here we say 0.20 2000 mole okay of hydrogen sulfide H2S times again we do our mole to mole ratio okay because we want to go exactly from the moles of H2S to the moles of H2O okay we are going from the moles of H2S to the moles of H2O so we say what is the molar ratio the mar ratio here is 2 is to2. Have you seen 2 moles of H2S will produce exactly 2 moles of H2O? So the molar ratio is 2 is to2. Then which one should be on top? The one we're looking for. We're looking for water, right? So we say for every 2 mole of H2O that is produced, two mole of H2S is consumed. And then this one here cancels with what we have below. This two cancels out the two. And then we have 0.200 mole of H2O * 1. Right. Correct. So we've canceled out the H2S. We now have H2O remaining.

[00:18:23]So this means that we have 0.2 mole of H2O.

[00:18:30]Okay. Now since we now have the moles of H2O, the next thing that we're going to do here is to convert from moles of H2O to mass of H2O. Okay? Because they actually ask us to find the amount in grams that can be produced. So what we do here we say remember or recall that the formula for moles is moles is equal to mass over molar mass which is written as mr. So when we cross multiply we now have that mass is equal to moles time mass which is also written as mr. So what is the moles that we have calculated? So our mass here is equal to moles = 0.20 multiplied by the mar mass. The mar mass of H2O is what? We need to also find find out. So here we have molar mass of H2O.

[00:19:37]So right here we have hydrogen is 1.008 008 right * 2 and then oxygen is equal to 16 right correct so when you add these two together you're going to get 18.02 g per mole so the molar mass for H2O is 18.02 2. Okay. So, you plug it in here and then you multiply to get the mass. We have 18.02.

[00:20:15]Now, the mass of H2O is equal to when you multiply these two, you're going to get 3.60 g of H2O.

[00:20:28]So, that is what's going to be produced given that exact amount of hydrogen sulfide. Okay. Yeah. So that is actually what it means in this very case. Now let's go over to the fourth one.

[00:20:43]The fourth one here says how many g of water sorry if 2 or 12.0 g of SO2 are formed.

[00:20:57]Let's be very careful with this very part. If 12.0 0 g of SO2 are formed from 7.98 g of H2S. What is the percent yield?

[00:21:14]If 12.0 g of SO2 are formed from 7.98 g of H2S, what is the percent yield? So, let me explain what percent yield is before we solve this very question. or at least it's going to give us an insight into what the question wants us to do. So right here percentage yield we say percentage yield is equal to actual yield.

[00:21:45]Actual yield. Okay. Over the yield.

[00:21:54]Theoretical yield.

[00:21:57]times 100%.

[00:22:00]So this actually tells you the amount that was collected or was produced after the reaction had gone into completion expressed in percentage. Okay. So let's assume that we solve this part now and then we get a 10%.

[00:22:22]What it means is that only 10% were produced. Okay? Or only 10% were collected. The orders, you know, were lost in the course of the reaction. So just like for example, let's assume you're writing an exam, right? The overall is 100 and then you get a 10.

[00:22:43]You score a 10 in the exam. So when you divide 10 by 100 you are going to get 10%. Okay that's when you multiply it by 100%. What this means is that you were only able to scoop 10% of the entire 100% possible.

[00:23:04]So here we have our actual yield already given in the question because the question here says if 12.0 g of SO2 are formed okay from 7.98 of H2S. So right here this one here is the product. This is a reactant. The reactants are producing something. Okay, we have two reactants here. We have H2S and O2. They are reacting to produce SO2, right? Correct.

[00:23:41]So, it says if exactly 7.98 g of H2S can produce only 12.0 g, now what is the percentage yield? So, here we have the actual yield. This was what was actually produced.

[00:23:59]Okay, this this is what was actually produced in the reaction. So now we don't have the maximum amount that could be produced assuming no losses. In order to help you understand this very concept better, let me use cooking to explain it. Okay, so we're going to be talking about actual yield and theoretical yield. So here, let's assume you're cooking, right?

[00:24:27]And then you start with about 10 g of water in your pot. Right? So you expecting that after everything after cooking everything you are supposed to have about let's say 9 g okay of let's say water or something like that.

[00:24:52]Now but after cooking you now observe that you got about about let's say 6 g of water remaining.

[00:25:02]Okay. So what this means is that this was the actual yield because when you start boiling water some of it will evaporate. Okay. So assuming there was no evaporation you would get 9 g.

[00:25:16]Okay. Let's assume that this was your expectation. We know that it was supposed to be 10 g. But let's just assume that this was your expectation.

[00:25:25]But now in the course of you know evaporation, heat and and all that, you lost a lot. Then you ended up getting about 6 g. Okay. So this one here is what was actually remaining while this one was what you were expecting. Okay.

[00:25:42]But then because of heat cause of evaporation cause some spilled you know in the course of cooking you you you lost about 3 g and ended up with with 6 g. So this one now became of it becomes rather your theoretical yield. Okay this now your theoretical yield. So in the same way we have our actual yield here which is 12.0 g but we don't have the theoretical yield cuz in the formula it says that the percentage yield is given as actual yield over theoretical yield time 100%.

[00:26:21]We have our actual but we don't have the theoretical. So what do we do in this very case? We need to calculate our theoretical yield. And how do we do that? How do we calculate our theoretical use? So, what we're going to do right here is that we have to use the amount of hydrogen sulfide that was reacted to do that and I'm going to explain to you why in the course of this solution. So, we have 7.98 g of hydrogen sulfide which reacted. So, we say 7.98 g of hydrogen sulfide.

[00:26:57]Now before we can do anything we need to convert from grams to moles. This is very important. Okay. So here we have moles of hydrogen sulfide contained in 7.98 g.

[00:27:13]Remember the formula is mass over molar mass.

[00:27:20]What's the mass here? The mass is 7.98 g of hydrogen sulfide. What is the mar mass? The mar mass is about 34.08. You can check that out at your free time.

[00:27:36]34.08 g per mole. So what to do here is you know that hydrogen is 1.008 and then you have about two of them.

[00:27:46]Right? Correct. Sulfur has an atomic mass unit of 32 32.06 right 32.06.

[00:28:01]So you just do the calculation and then here we have moles of hydrogen sulfide contained in 7.98 g is equal to exactly.

[00:28:15]Let's do the division here. So, we're going to have 0.234 0.234 moles of hydrogen sulfide H2S.

[00:28:28]Okay. Now, having found this, we now have to go back to the balanced chemical equation. Okay. So, here we have 0.234 mole of hydrogen sulfide.

[00:28:43]Okay. So here if you look at this chemical reaction here. Okay. So this stochometric coefficients actually tell you the maximum amount possible that can be produced under ideal conditions. When we say ideal conditions we mean there are no losses. Okay. So with these numbers in front here you can easily find your theoretical yield. Okay. So these numbers actually says for every two moles of H2S that reacts two moles of SO2 will be formed but it is only if there are no losses. Okay. So these are under ideal situations where there are no losses, there is no heat, there is nothing. So everything will be produced all the reactants will be converted into what our product. So that is actually an ideal situation. So we use that to calculate the theoretical yield. Okay.

[00:29:45]So we say what is the mole to mole ratio. So for the mole to mo ratio here we have that for every two moles of SO2 that was produced we have that two mole of H2S was consumed. You can go back and see it there. Okay. So, right here, for every two moles of SO2 that was produced, about two moles of H of H2S were consumed. Okay. So, let me say that again so that you understand. So, for every two moles of SO2 that were produced in a reaction, two moles of H2S were consumed. Okay. So doing our mole to mole ratio right now we're actually going from the moles of the reactant to the moles of the product. So what this means is that for every two moles of H2S that is consumed we have two moles of SO2 that we produce. So we are actually looking for moles of SO2 right not H2S.

[00:30:56]So this H2S here cancels with H2S down here. So 2 cancels with 2. We now have 0.234 mole of SO2. Right? Correct. Cuz this is now a 1. 2 into 2 is a 1. So 0.234 * 1 will give you just the same value. Okay.

[00:31:19]Now having calculated the moles of SO2 that can be theoretically produced that is a theoretical yield. We now need to convert this to grams. Okay? Because in the formula we don't use moles. In this formula we use g. So the actual must be in g. Theoretical must also be in g. And then multiply by 100%. So we have to convert from moles of SO2 to g of SO2.

[00:31:50]So we say the formula again we have moles is equal to mass over mass. Okay.

[00:31:58]So we cross multiply mass * 1 is mass and then moles time mass that will give you moles time mass. Then we can now plug in to solve for the mass. So the moles that we calculated were 0.234 right and then the mar mass of SO2 you can check that that will give you 64.06.

[00:32:25]Okay. So the mar mass of SO2 is equal to 64.06 64.06 g per mole. Please look that out.

[00:32:38]You can find it. So we have 64.06.

[00:32:43]Now when [snorts] you multiply these two, we're going to get 15.0.

[00:32:49]So the mass of SO2 that can be possibly produced in ideal situations equals 15.00 g okay of SO2. [snorts] So this is our theoretical yield. Now we can now go ahead and calculate to see what we're going to find. So we say um percentage yield percentage yield is equal to actual yield over theoretical yield theoretical yield times 100%.

[00:33:37]So our actual yield according to the question we were given 12.0 right correct. So we have 12.0.

[00:33:47]So this is equal to 12.0 g remember they must be in g over theoretical yield is 15.0 g* 100%. So g cancels g. So this is equal to 80%.

[00:34:06]when you do the calculation you're going to get 80%. So which means that um 20% were lost okay due to heat due to vibration and other situations. So this is the percentage yield 80%.

[00:34:21]Now let's solve the other two parts. So the last two says how many g of SO2 were produced starting from 2.66 g and 3.00 00 g of O2. How many g of SO2 were produced? So right here we have this react with this right to form SO2.

[00:34:47]So they're asking us if a certain amount of H2S okay reacts how many or what what amount exactly of SO2 will be produced again if a certain amount in grams of oxygen gas reacts how many amount or what amount exactly of SO2 will be formed that's what the question is trying to uh ask us. So here what we need to do right now is to always understand that whenever you want to go from the uh from the grams of one specy to the grams of another you need to first convert to moles then you do your mole to mole ratio. Okay don't forget that. So right here we have we're given 2.66 g of H2S. So first thing to do here is to convert g to moles. Okay. So let me do that right here.

[00:35:50]So we have 2.66 g of H2S.

[00:35:56]So this is uh the fifth one. Is it the fifth one?

[00:36:00]Okay. Yeah. So we have 2.66 g 2.66 g of H2S. Right. Okay. Correct. So we now want to convert from grams to moles. We say moles of H2S equals remember the formula is mass over molar mass.

[00:36:29]So moles of H2S is equal to the mass we're given is 2.66 right 2.66 66 g over the molar mass as we initially calculated was 34.08.

[00:36:49]So we have 34.08 g per mole.

[00:36:54]Okay. So when you do this division we are going to get that the moles of H2S moles of H2S is equal to 0 0 780 moles.

[00:37:17]Okay. So that is it. And now we can also calculate the moles of O2. So let's see how we can do that.

[00:37:34]So we say moles of O2.

[00:37:39]We were given exactly 3.00 g, right? So we're given 3.00 g. We say given 3.0 00 g of O2. So what we do right here is that we also set up the same formula. So we say moles of O2 is equal to remember that the formula is mass over mass.

[00:38:02]Right here we have 3.00 g of O2 over the molar mass of oxygen is 32, right? Cuz we have 16 for one. If you multiply that by two, that give you 32 g per mole. So divided by 32 g per mole. So this is equal to when you divide 3.0 by 32, the answer is 0937.

[00:38:31]0.0 937 mole of O2. Okay. So having found their respective moles, we can now do a mole to mole ratio. Okay. to know exactly the amount of SO2 that can be formed given the amount of our reactant.

[00:38:57]So we start with the first one. So H2S to S O2.

[00:39:06]Okay. So here we have 0.0780 0780 mole we say we have 0.0780 mole of H2S. Okay. So using the same chain link method we can do our more to mole ratio.

[00:39:29]The question is which one are we looking for? Are we looking for SO2 or H2S?

[00:39:34]We're looking for SO2. Right. Correct.

[00:39:37]So we do our mole ratio. the one we're looking for has to be on top. Now, what is the mole to mole ratio here between H2S and SO2? So, going back again to the balanced chemical equation, we have that for every two moles of H2S that reacts, two moles of H of SO2 are formed. So, we represent that we say we're looking for SO2. It has to be on top. For every two mole of SO2 that is produced about two mole of H2S are consumed. So mole of H2S cancels with mole of H2S and then 2 / 2 that give you a one right. So 0.0 780 * 1 will give us 0.0780 0780 mole of SO2.

[00:40:34]Okay. Now we have moles of SO2 produced from um moles of H2S. Now let's do again we do oxygen to SO2. That's the next one.

[00:40:51]So oxygen to S2 we have O2 to S O2. So we have exactly exactly 0.0937 mole of O2.

[00:41:09]We have 0.0937 mole of O2. So we say times again we do our mole to mole ratio. Remember you do your mole to mole ratio when you want to go from the mole of one reacting specy to the mole of another. Okay? You cannot just jump from mole to mo from speci to specy. You have to exactly uh go through the bridge and that bridge is what we call the mole to mo ratio. The secret is that the one you are looking for has to be on top and the one you want to cancel out has to be below. Don't forget. So here we go back again to the mole to mole ratio to the molar ratio. So looking in front of this reacting species here for every three moles of oxygen that reacts two moles of SO2 are produced.

[00:42:12]So, which one are we looking for? SO2, right? It has to be on top. So, right here we have for every two moles of SO2 that is produced, three mole of oxygen is consumed. So, the mole of oxygen cancels with the mole of oxygen. And then we have 0.0937 0 937.

[00:42:42]Okay. Time 2 mole of SO2 all over 3.

[00:42:50]Right.

[00:42:51]Correct. All over three. So when you multiply these two together, let me clean this very part so that we can continue from down here. So when you multiply these two together you are going to have 0 62 46 mole of SO2.

[00:43:21]Okay. So right now we have the exact amount that can be produced. Okay. But we still have them in moles. So we need to convert from moles now to gram because the question says how many g of SO2 are produced from these two reactants. So now we have gotten the exact amount of moles that were produced. We now have to convert again from moles to grams. So let's see how we can do that very quickly.

[00:43:55]right here we shall continue.

[00:44:00]So let me clean this very part.

[00:44:06]Okay. So here we say again um grams or we say moles to grams.

[00:44:18]Moles to grams.

[00:44:24]moles to grams. So we see how we can do the conversion. So right here we have 0 0 780. Okay. So 0.7 0.0780.

[00:44:48]So over here we have moles [snorts] is equal to mass over molar mass. Okay.

[00:44:55]So we cross multiply.

[00:44:57]We have that mass is equal to moles.

[00:45:02]Okay multiplied by molar mass.

[00:45:08]Now we start with the first one.

[00:45:10]So the mass of SO2 produced by H2S is equal to so we had the mass is 0.0 780 times the molar mass of um SO2 right now. Okay cuz we're actually going from moles of SO2 to mass of SO2.

[00:45:37]So the mar mass of SO2 is 64.06.

[00:45:43]So when we do the calculation, we're going to get 5.00 g. 5.00 g of S2.

[00:45:56]Okay. So this actually implies that 2.66 g of H2S can produce 5.00 00 g of SO2.

[00:46:09]That is what it means. So let's now do the next one. [snorts] Let's use what oxygen produced of SO2 to check. So we can now go again from moles to g or to mass. So again we say mass is equal to moles time mar mass.

[00:46:35]So the mass of SO2 that was produced from exactly 3.00 g of oxygen is going to be right here. We got it to be let's go back before I forget 0.0624 right correct so we have 0.0 624 okay multiplied by 646 multiplied by the molar mass of um SO2 which is 64.06 06 64.06.

[00:47:20]So when you do this calculation, what is the answer?

[00:47:25]So this is our calculator check. So right here we have 0.0 624 6 * 64.06.

[00:47:38]So I'm getting 4.00. So this is equal to 4.0 00 g of S O2. So let me explain to you what this physically means. So this means that 2.66 g of H2S was able to to produce 5.00 g of SO2 and again 3.00 00 g of oxygen was able to produce only 4.00 g of SO2. Okay. So we have answered that very part. Now the last part says what is what reactant is a limiting reagent?

[00:48:27]Okay. Or which reactant is a limiting reagent?

[00:48:32]Okay. So now how do we answer this very question? So by definition um a limiting reagent is the speci in the chemical reaction that is wholly wh l y used up in a reaction. Okay. So it can also be said to be the one that produces the least amount of the product. It can also be said to be the one that finishes first in a chemical reaction.

[00:49:12]Okay? So when you have two reactants reacting to form a product, the one that produces the least amount amongst the two reactants is the limiting reagent.

[00:49:24]If you observe very closely right here, we have that um H2S was able to produce 5.00 g of SO2 while oxygen was only um able to produce 4.00 g of SO2. So which one produced the least amongst the two? It was oxygen, right? Correct. So oxygen produce the least amongst the two.

[00:49:55]Therefore oxygen is the limiting re agent. Okay. So I say again the limiting reagent is the one that produce the least amount of the product or it is the one that finishes first in a chemical reaction. So oxygen got to 4.00 00 g of O2 of SO2 and stopped. But then you can see that this one is still ahead. This one is still bigger. So this one here is in excess, but oxygen finished first or it was able to produce the least amount of SO2. Therefore, it is a limiting reagent. In case you have any questions, comments or concerns, you can reach out to me in the comment section and I'll be right there waiting for you. Thank you and see you in the next video.