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## Stoichiometry

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# Stoichiometry example problem 1

## Video transcript

- [Voiceover] We know that solid phosphorus will react with chlorine gas to spontaneously, to produce phosphorus trichloride, liquid phosphorus trichloride, and we're told that we have 1.45 grams of solid molecular phosphorus and we're asked, "How many grams of chlorine is required "to essentially use up all of the phosphorus that we have, "and how many grams of phosphorus trichloride "is going to be produced?" Now before you do any of these stoichiometry problems, and that's just a fancy word for problems where you need to figure out how much of a certain reactant is required, or how much of a product is going to be produced. Before you do any of these problems, you have to be sure that your reaction, or that your equation is balanced. So let's make sure. So on the left-hand side here, this molecule of phosphorus has four phosphorus atoms, so on the whole left-hand side are, all of our reactants combined have four phosphorus atoms. So our products need to also have four phosphorus atoms, but the way it's written right now, we only have one. So let me just multiply this guy by four. Now I have four phosphorus atoms on both sides of the equation. Let's balance the chlorine now. On the left-hand side I only have two chlorine atoms. This one molecule of chlorine has two atoms in it. Here I have, each molecule of phosphorus trichloride has three chlorines, and I have four molecules of it, so four times three, I have 12 chlorines on the right-hand side. So I have 12 on the right-hand side, I need to have 12 on the left, I only have two here. Let me multiply that by six. Six times two is 12, four times three is 12. Now our equation is all balanced. Four phosphoruses on each side, and 12 chlorines. Now the next thing we have to do, now that we know that we have a balanced equation, and we can kind of get into the meat of the problem, is figure out how many moles of phosphorus we're dealing with. Because once we know the moles, we can use the stoichiometric ratios, which is just essentially saying, "Look, "for every mole of that, I need six moles of that, "and for every mole of that, "I'm gonna produce four moles of that." So you want to get it all in terms of moles. So let's figure out how many moles of phosphorus we have on our hands. And let's look at our periodic table. This periodic table, I have to give proper attribution to the maker. It's made by Levon Han Cedric, I got this off of Wikimedia Creative Commons, it had an attribution license. Other than that, we're free to use it. And lets go to phosphorus. Phosphorus right here, phosphorus right here has its atomic weight of 30.974. Let's just round that up to 31. Let me write it down right here. So phosphorus, phosphorus has an atomic weight atomic weight, of 31, which tells us that a mole of phosphorus will weigh 31 grams. Remember a mole is this, you know, 6.02 times 10 to the 20th, this huge number of atoms. If you have that many number of atoms of atomic phosphorus, it's going to weigh 31 grams. Now if you look at the atomic weight of P four, or a molecule that has four phosphorus atoms in it it's gonna be four times this. So it's going to have an atomic atomic weight of, well what's four times 31? It's 124, of 124, which means that one mole of, let me write it here. This tells us right there, that one mole of solid molecular phosphorus is going to have a well, since we're doing grams, I should say, will have a mass of 124 grams. Now, given that, we can use that with this information to figure out how many moles of molecular phosphorus we have. Solid molecular phosphorus. So let me start over here. So we have 1.45 grams, let me write it this way, grams of phosphorus, of molecular phosphorus, each molecule containing four atoms, and we can just do a little dimensional analysis, make sure everything cancels out, times this information right here. We have one mole of phosphorus, of molecular phosphorus for every 124 grams of molecular phosphorus. I should write this here, just so we remember what we're talking about. 120, one mole of molecular phos for every 124 grams of molecular phosphorus. And then this cancels out. So what we have, the units at least cancel out. The grams of phosphorus, grams of phosphorus, and then we get 1.45 times, what do we have, times one over 24, times one over 124 moles, moles of molecular phosphorus. And we could figure that out. Is equal to 0.0 1-1-7 moles of phosphorus. That's what we're starting off with. That' s what this 1.45 grams are. And that makes sense. If we had an entire mole of molecular phosphorus it would weigh 124 grams. We only have 1.45 of that, so it's almost, you know, it's a little bit more than one hundredth, which makes sense. This number right here is a little more than one hundredth. Now we need to think about for every mole, let's scroll down here, although I don't want to lose my equation. We need to think about for every mole, so let me write down what we just, so we have 0.0117 moles of molecular phosphorus, and for every mole of phosphorus, how many moles of chlorine molecules do we need? So let's, let me write this down. So for every, and I'll write it this way. For every six moles of chlorine gas, I'll do it in blue. So times, for every six moles of chlorine gas, we need one mole, we need one mole of molecular phosphorus. And the reason I wrote it this way instead of writing one mole of molecular phosphorus for every six moles of chlorine gas, is because I want to make sure that the units cancel out. And it also should make sense for you intuitively. If I have, if I have one of this, I'm gonna need six of this. If I have 0.0117 of this, I'm gonna need six times of the chlorine gas, and this, the units work out. This cancels out with that, and I'm just essentially multiplying by six. So we're going to have six times, let me multiply it, six time 0.0117, so this is equal to, 0.07 moles of chlorine gas required. And I could write it right here. Let me write "required." Required in the numerator, required in the denominator. That makes a little bit more sense or it clarifies things. For every six moles of chlorine gas that are required, one mole of phosphorus, of solid phosphorus is required. Or you could say for every mole of phosphorus is required, you need six moles of chlorine gas. We got that just from the original equation. So I'll write this "required" down right there. So we're almost there. We figured out that 0.07 moles of chlorine gas are required. But what we want to find out is how many grams of chlorine gas are required, so we just have to figure out how many grams there are per mole. So let's figure that out. Let's look at chlorine. Chlorine is off here to the right. This is a super huge periodic table. Chlorine right here has an atomic weight of 35.453. So chlorine, chlorine, 35.453, atomic weight. The weighted average of all of the isotopes of chlorine on earth. So chlorine, molecular chlorine gas, which has two atoms, is going to have twice that, so I could do it in my head, it's gonna be right about what, 70, 70.906, atomic weight, which tells us that, well let me write this information that we had down before, we are required to have .07 moles of chlorine gas, that is what is required, and let's multiply it. We want moles in the denominator. So we want moles in the denominator here, so one mole of chlorine, of chlorine gas, for every mole, how much is that? How much is that, what's the mass going to be? Well if the atomic weight of chlorine gas is 70.906, that means one mole of it is going to have a mass of 70.906 grams. So for every mole, we have 70.90 grams, we have .07 moles, so we'll multiply .07 times 70 to figure out how many grams we have. And the units cancel out. We have moles of chlorine gas, moles of chlorine gas, and we're just gonna have grams required. So this is going to be equal to 70 times .07, or .96 grams of, and actually, this is grams of CL, of chlorine gas, I should write that there, so grams of chlorine gas are required, even the required worked out with the dimension analysis, and we've answered the first part of the problem. If we have 1.45 grams of phosphorus as one of the reactants, then we're gonna need 4.96, 4.96 grams of chlorine are required. Now the second question is how much phosphorus trichloride is produced. Now there's two ways to do it, an easy way and a hard way. The easy way says, "Well, mass is conserved." You know, if you have grams on this side, the total mass on this side has to be equal to the total mass on that side. That's the easy way. The slightly harder way is, you could have done the exact same thing we did to the chlorine gas, you could do it with the phosphorus trichloride. You could say, "Hey, for every mole of this, "I need four moles of that," and then figure out the mass of each mole and multiply and do all of that. But let's do it the easy way. We know that we have 1.45 grams of this reactant, and we have 4.96 grams of this reactant, so this, which is the only product right here, this product right here, that must have a mass of our combined reactants, 'cause it's going to react fully.