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## Stoichiometry

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# Limiting reactant example problem 1 edited

## Video transcript

- We're told methanol, which is used as a fuel in racing cars and fuel cells, can be made by the reaction of carbon monoxide and hydrogen. So, this is the methanol right there. They're giving us 356 grams of carbon monoxide, so carbon monoxide, we have 356 grams of it and they're giving us 65 grams of hydrogen, of molecular hydrogen, 65 grams. They're mixed and allowed to react and they say, what mass of methanol can be produced? And then, they say, what mass of the excess reactant remains after the limiting reactant has been consumed? So, that tells you that this is a limiting reactant problem, that we have too much or too little of one of these two reactants. These are the two reactants there. The one that we have less of is the limiting reactant, that'll dictate how much of the product we can produce. And, the one that we have more of is the excess reactant. But first, we have to figure out which is the limiting and which is the excess. And, before we even do that, we should always check that our equation is actually balanced. So, let's just check that. On the left hand side of this equation, we have one carbon, right there. On the left hand, on the right hand side, we also have one carbon, so that looks good so far. On the left hand side of the equation, we have one oxygen and on the right hand side, we have one oxygen. Looks good so far, left hand side we have four hydrogens, two times two, on the right hand side we have four hydrogens, three plus one, so this is balanced. It is balanced, so we can proceed to try to figure out what the limiting reactant is. So, the way we wanna do it, is figure out how many moles of each of these were given and then figure out how many moles, the stoichiometric ratio that's required by this reaction. We already know what it is, we know that for every two moles two moles of hydrogen required, we require, you can see it right here from the equation, one mole of carbon monoxide, one mole of carbon monoxide required. This is what the equation tells us, now let's see how many moles of hydrogen and how many moles of carbon monoxide we have. The first place to start, we've done this several times, is what is the molecular weight of carbon monoxide? So, carbon monoxide molecular weight, let me write it here, molecular weight for carbon monoxide. It's a carbon, has a molecular weight of 12. It's good to memorize that, carbons and oxygens and hydrogens show up so frequently. And then, oxygen is 16, oxygen is 16, so it's equal to 28, molecular weight of 28, which tells us that carbon monox-, so, if we wanna do it this way, so let's do the carbon monoxide first. So, we're dealing with the carbon monoxide. So, we have 356 grams of carbon monoxide and we wanna write this in terms of how many moles of carbon monoxide do we have. So, what we want is how many, how many grams per mole, but I should write this, I guess one mole has how many grams? So, one mole of carbon monoxide has how many grams? Well, it's molecular weight is 28, so, if we have mole of them, if we have that, you know, 6.022 times 10 to the 23rd molecules of carbon monoxide, that's going to weigh 28 grams. So, one mole of carbon monoxide is 28 grams or I guess, yeah one mole per every 28 grams and the reason why I wanted to put the grams in the denominator is so it cancels out over here. So, those cancel out when we multiply and so, we are left with 356 times one, divided by 28 is 12.7 moles of carbon monoxide. Now, let's do the same thing for the hydrogen. Let's do the same thing for the hydrogen. We have 65 grams of molecular hydrogen. What's the molecular weight of hydrogen? I'll do this in green over here, molecular weight of hydrogen is, well each hydrogen atom has a molecular weight of one times two, which is equal to two. So, we have 65 grams of molecular hydrogen and the same way, we wanna write it in moles, so we're gonna say, we're gonna multiply it times one mole of hydrogen is equal to how many grams? Well, we just figured out one mole is equal to it's molecular weight is two, so a mole of it is going to have a mass of two grams. So, you can view this as two grams per mole or one mole per two grams, and we want the grams in the denominator so it cancels out over here. And so, let's do the math, that cancels out with that and we have 65 times one, divided by two. 65 divided by two is what? 32.5, 32.5, 32.5 moles of hydrogen, of hydrogen, moles of hydrogen. Now, we know exactly how many moles of carbon monoxide and how many moles of hydrogen they've given us. Let's figure out what the ratio is and we'll do it in the exact same way as we wrote up here. So, we have 32.5, this is what we're given, moles of hydrogen, let me write that a little bit neater. Moles of hydrogen and we're given 12.7 moles, I'll do that in the same color, 12.7 moles of carbon monoxide. So, if we were to just divide this, what is this? I guess you could imagine, divide the numerator and the denominator by 12.7. So, this can be rewritten as, so if I just rewrite this, I should've written it here to begin with. I could write this as we have 2.56 moles of molecular hydrogen for every one mole of carbon monoxide. For every one mole of carbon monoxide. So, what, this is what we need for a reaction to occur. This is what the balanced equation tells us, it tells us that we need two moles of hydrogen for every mole of carbon dioxide. Based on what they've given us, we just figured out that we have 2.56 moles of hydrogen for every mole of carbon dioxide. So, we have more than enough hydrogen, right? We only need two for every mole of carbon dioxide, we have 2.56. So, we have an excessive amount of hydrogen. So, the excess reactant is the hydrogen. Hydrogen is excess, excess reactant and the other one is going to be the limiting reactant. We don't have enough carbon monoxide to react all of the hydrogen, right? We only have one for every 2.56 we would actually for this says, you need one point, you know, two five or whatever for every, or 1.28 for every 2.56. It should be a one to two ratio, so we don't have enough carbon monoxide to react all of the hydrogen. So, carbon monoxide, carbon monoxide is limiting, limiting reactant. Now, given that this is the excess reactant, let's see how this, the, we can use stoichometric ratios to figure out how much methanol's going to be produced. It's all going to be limited by our, by our carbon monoxide. Did I just say that this, hydrogen's not the limiting reactant, carbon monoxide is the limiting reactant, we have more than enough hydrogen. So, how much carbon monoxide do we have? We already figured it out, we have 12.7 moles of carbon monoxide and we look at our, we look at our, let me write this over here. So, we have 12.7 moles of carbon monoxide and looking at our original equation, we see for every mole of carbon monoxide, we produce one mole of methanol, let's write that down, times, and we want the carbon monoxide in the denominator. So, for every one mole of carbon monoxide used, we have one mole of methanol, which is what? CH3OH, did I get that right? Yep, produced. We get that straight from the balanced equation. And, this math is pretty easy, but it gives us the right unit. If we're gonna be, remember, we're using the carbon monoxide, not the hydrogen, 'cause the carbon monoxide is the limiting re-agent. That's what's telling us what's going, you know, if we can't, if we used hydrogen as the limiting reactant, then we wouldn't have enough carbon monoxide for the reaction to occur, so we're... This is what's kind of capping off on how much this reaction can move forward. But, the whole point of this was to cancel that and that. So, obviously if we're using 12.7 moles of carbon monoxide, we're going to produce 12.7 moles, 12.7 moles of methanol will be produced. Will be produced and now, we just have to figure out how mu- what is the mass of 12.7 moles of methanol. And we just think about what the atomic weight, so, if you look at methanol, CH3, let me put the H3 there, H3OH, it's atomic weight is 12 plus three times one plus three plus 16 plus one, so what is this? This is 20 plus two is equal to 32 or we could say that, if we'd think in molar terms, or not molar terms, I should just say 32 moles, this is it's atomic, this is it's atomic weight, sorry atomic weight, so that tells us that if we have a mole of it, we're gonna have 32 grams of methanol per one mole, per one mole of methanol. And once again, we got that by figuring out it's atomic weight. Now, to convert the number of moles we have of methanol to the number of grams, we just multiply that times that, the units work out, right, this is in the numerator this is in the denominator, let me just copy and paste it. So, we have that, copy and paste, and then, we can multiply it times that. Let me copy and paste it, get that right there, and I'll pick a different color, maybe a blue, and just like that, and then, let's multiply these two. We have moles of methanol, moles of methanol moles of methanol and we're left with 12.7 times 32 is equal to four o, actually I should just stay with (illegible), so we'll say 406 grams, I'm rounding down, 406 grams of methanol, let me write it with the units. So, grams, I could, grams of methanol, went off the screen, of methanol produced, I could write the produced down here since we're all done. So, I think that was the first part of our question, how many grams of methanol, what mass of methanol, that's 400 and, I have a horrible memory, 406, 406 grams. And then they say, what mass of excess reactant remains after the limiting reactant has been consumed? Now, there's a couple of ways you can do this. The easiest way, the easiest way to think about it is that the mass has to be conserved. So, we started off with, we started off over here, with 65 grams of, let me be careful. We started off here with 356 grams of carbon monoxide and 65 grams of hydrogen and we were able to produce 406 grams of methanol, 406 grams, and we figured out that carbon monoxide was the limiting reactant. So, all of this, all of this gets consumed and only some of this gets consumed. So, if you do the math, what gets consumed what gets consumed has to be equal to 406 grams, 'cause that's what gets produced. So, let's think about it a little bit. The left hand side, how many total grams do we have? So, if we have 356 plus 65, we're starting off with 421 grams of reactant. So, we're starting with 421 grams of reactant, 421 grams of the reactant, and then we end up with 406 grams 406 grams of product, of our methanol. So, that means that 421 minus 406 grams of reactant was not used, so that means that 421 minus 406, which is equal to what, 21 minus six is 15 grams of reactant not, not used. Now, the reactant that's not going to be used, is the one that you have an excess of and we figured it out that we have an excess of hydrogen. We have an excess of hydrogen, so all of that 15 grams must have been 15 grams of hydrogen that was not used. So, 15 grams of hydrogen left over. Now, the other way you could've done that, the other way you could've done this exact same problem, is, you could've said, look we're starting with 12.7 moles of carbon monoxide, that's the limiting reactant. It's a two to one ratio, you two need moles of hydrogen for every mole of carbon monoxide. So, you said, okay we have 12.7 moles of this, I need twice that many moles of hydrogen. So, you would say, well what's twice that, that's 20, that's 25.4. You need 25.4 moles, you have 32.5, so you'd subtract the difference. You'd subtract 32.5 minus 25.4 and that number of moles of hydrogen is left over. You'd multiply it times the grams per mole, which is two and then, you, once again, would get the same thing. You would get 15 grams, let's do that, I just wanna make sure you understand. So, we're starting off with, we are starting off with 12.7, 12.7 moles of carbon monoxide and we know, and we know that we are required, that two moles of hydrogen are required for every one mole of CO, of carbon monoxide, that's the required, that is required. So, you know that this cancels with that. You know 12.7 times two is 25.4 moles of H2 required and what is the mass of 25.4 moles of H2? So, let's write that, 25.4 moles of hydrogen required times, well we know the molecular weight of H2 is two, so one mole, we're gonna have two grams of H2 per one mole of H2. This cancels out mole of H2, mole of H2, so we're gonna have 25.4 times two is what, that is 50.8, that is going to be 50.8 grams of H2, of H2 required, required, that's how much we're going to consume in the reaction. Now, they wanted to know how much is left over. So, we're going to consume 50.8, we started with 65. So, if you subtract 50.8 from 65, 65 minus 50.8, you're going to get what? You're gonna get 14 point, you're going to get 14.2 grams left over and that compares, that compares to the 16 we got left over, when we just took the 421 minus the 406. And, the difference between the 14.2 and the 15 is really just a little rounding here and there with our digits. Anyway, hopefully you found that helpful.