 So, the problem that we're going to do is ethyl alcohol is prepared industrial by the reaction of ethylene C2H4 with water, H2O. What is the percent yield of the reaction if 4.6 grams of ethylene is 4.7 grams of ethyl alcohol plus 1. So let's go ahead and do this. So it gives you the reaction equation, this is C2H4. So it also gives us the mass of ethylene, C2H4, 4.6 grams, and it also gives us the mass of ethyl alcohol as 4.7. So what's the percent yield? Do we just say 4.6 divided by 4.7 and knock it out? No. We've got to do what? Convert it to moles. Convert it to moles. And this reaction, hopefully you guys are asking, well, how do I know which one is the limiting reagent? Okay? Hopefully that was the question you're asking yourself because they don't give you the mass of water, right? Remember what we said, well, if they don't give, first of all, if they don't give you the mass, then they must be saying the other things, the limiting reagent, okay? So in this case, ethylene is, but not to mention the fact that water is very inexpensive. Remember when we were talking about limiting reagents? Limiting reagents are usually the more expensive or the less prominent or predominant thing. So ethylene, of course, is going to be the limiting reagent. If they don't, obviously I don't expect you guys to always get that, you know? But if they don't give it to you, then you just do assume that that one is the limiting reagent. Okay? So let's go ahead and figure this out. So we've got to figure out, well, what's the molar mass of both of these things? Okay? Ethylene is going to be 2 times 12.01 plus 4. So molar mass of ethylene is 28.052 grams per one mole, and this is C2H4. So that's going to give us the number of moles of ethylene. Yeah, let's go out to a few digits for ethylene. 6, well, we'll just say 6 more. We'll worry about our sick things at the very end, okay? And now we've got to do the same for ethanol. So we know one mole, 0, 6, 8 grams of ethanol. The number of moles that I got of ethanol are 0.102 moles of C2H6. Okay? So now we're comparing what is the liquid, the liquid with the product. We've figured out what the percent yield is. Is everybody okay with that? So figure out the percent yield. The number of moles of limiting reagent divided by the number of moles product divided by the number of moles limiting reagent. So first, yeah, 1, 1, 2, moles divided by 4, moles times 100 percent. Notice the moles cancel, 0.2 percent. But when we come back over here, it's 2, 6, 8. So let's just take it to 2, 6, 6. So it's going to be 62 percent yield. I think it's a 0.164 mole of C2H4 of limiting reagent. Pardon? Why was that top 1 of limiting reagent taken from that first? Because this is the reagent in excess. We will. Yeah. So, I mean, it don't give you any information about it. So, I mean, this, what are you saying? Why isn't this it? Yeah. This is the product. This is not something that you're, this is not a reactant. This is a problem. I think you're understanding this. So the thing is, is the limiting reagent is going to be one of the reactants here, you know? This is the product. Okay. So you can only, it's like the race. Remember we were talking about the race? So the limiting reagent of a race would be how many people started it, okay? So, and we're talking about starters, right? It could be only this thing or this thing, right? This is like how many finished it, okay? So if we had 164 that started it and 102 that finished it, that's what we're getting our percent from. Okay. Does that make more sense? Yeah. Okay, cool. Just think about it like a race, okay? Yeah. Like multiple reactants, so you have to actually figure out. You have to figure out what the limiting reagent is, right? So if in this case it said the mass of water is, I don't know, 18.02 grams, right? Then you'd have to use that to see if that's the limiting reagent. Like would it be the one with the least amount of moles? The least amount of moles. So in this case, this, right, moles of H2O, right? So I compare that to that, which one's less? The echo one, right? So that's still the limiting reagent, okay? Does that answer your question? Does that answer your question too? Make more sense now? Okay, good. Any other questions about that? So remember, you've got to do the limiting reagent first if they're both given to you.