 Okay, so let's do this first problem that doesn't have an answer to it already. So in an experiment to prepare bromobenzene according to the equation. So let's draw the equation on the board. So it's C6H6 plus Br2 goes to C6H5Br. Br plus each Br, right? So this is bromobenzene right here. So in an experiment to prepare bromobenzene according to that equation a student reacted 20 grams of benzene which is C6H6 with 0.310 moles of bromine. If 28 grams of bromobenzene was obtained what was the percent yield? Okay, so this is a pretty involved problem. Now that I look at it. So let's go ahead and do all the parts to it. Okay, so let's go ahead and write out what it gives us first. So it says the number of moles of bromine and it gives us the mass. And it also gives us the mass of bromobenzene that was obtained which is 28. So the first thing we're going to have to do is, well we're going to have to convert both of these things to moles eventually, right? But once we do that we're going to compare this one and this one to see which one is the limiting reagent, okay? And then we're going to compare the number of moles of that to the number of moles of this and see the percent yield, okay? So are we all cool with that? You guys think you can do this by yourself all that, right? Give me at least a, okay, yeah, at least some of you can. Okay, so first thing we got to do is find the molar mass, right? So remember from grain to moles of 6 times 12.016 times 1.0078 moles of C6H6 or one mole of C6H6, right? So that will cancel that out and that should give us the number of moles of benzene. So 20 divided by that answer 78.1, whatever. So 0.256, we'll just go to moles of C6H6. Okay, so now we've got a mole to mole comparison, so that's numbers, right? So that's actual numbers of things, right? So we can actually compare which one we have more of, right? Relative to the chemical equation. So here we've got a one to one mole ratio of bromine to benzene. So, I mean, this one's pretty straightforward, but what we could do is say, well, how much bromine would we need if we had that much benzene? We could take it one step further and let's just do that just for fun. Or we could do it the other way. How much benzene would we need if we had this one's bromine? Let's do it that way, just so we don't have all that string on the other side. So what's the ratio here? It's a one to one ratio, right? So one mole Br2 to one mole C6H6. So that cancels that out. And then, of course, that gives us, so we could compare these two numbers. So hopefully you see, right, that the benzene is the limiting range, right? Why? Because it's a less amount, okay? So now let's go ahead and do the number of moles of bromobenzene. So 12.01 times 6, 35, 79.90. And I get 157. So if anybody got anything different from that, 28 divided by 157. And that gives us the number of moles of product. So we got 178 moles of C6H5P. Okay, so remember this was the limiting reagent. So we're going to compare this to that. Okay, because this is the maximum amount that could have been made, right? Why? Because we have a one to one ratio of the limiting reagent to bromobenzene. Okay, so let's just do the problem then. So the percent yield is going to be the total number of moles that you could have had. So if you want to write it out, total. And this would be of the limiting reagent if you want to even put that. And the top would be the number of moles of product. And in this case, the product of bromobenzene. You're going to multiply that by 100%. Notice moles is going to cancel out here. The units are going to be percentage. So let's go ahead and do it. 0.178 moles divided by 0.256 moles times 100%. Cancel, cancel. Okay, so divide that by... I got, correct me if I'm wrong, 69. So what do we start with? 3. So 69.7%. That's 5. That's because I kept all of those digits there. That's okay. You're not supposed to round until the end anyway. If you start rounding it before you're going to introduce a lot of errors. But anyways, 69.7, 69.5, same thing. So you look at multiple choice tests. Okay, so any questions on that? You guys could have probably done that on your own, right? Yes. Okay, cool. Question before I turn it off? Yeah. Okay, why did on the top one, why didn't you do it to C6H6? I'm going to just compare it straight away. Okay. So I mean, the difference, I mean, why would you do that? Because what if this was a 1 to 2 ratio? Then you're going to want to figure out, you know, what's the mole to mole ratio of this to this that you've got? Okay. And you always did the limiting reaction on the bottom? It's always the limiting reagent because that's how many you've got. It's like this. If, you know, 30 people show up, then the maximum amount of days I could give is 30, right? It's a good analogy for students, man. They like it. Any other questions on it? Okay. Maybe we'll do another one where we find, we have a difference in moles to moles so we can actually see more of what that question that Peggy was talking about.