Okay, so we this is a percent yield problem. And ah, we are given the combustion of octane once again, and we're told that we observe the formation of 1.90 times 10 to the third grams of co two. Um, we have to figure out the percent yields of this reaction. And how many grams, um, are possible to be formed, Right? So that's that's the equivalent to the, um, the theoretical yield of this reaction. So we're gonna start with one leader of octane and we're gonna have to get to grams so we can get two moles, right? So the density of octane is 20.700 grams per millimeter. So let's first convert believers, the middle leaders using the prefix from an earlier chapter. Right? So if every 1000 milliliters is one leader or, um, another way to express that is for everyone. Mill leader. There's 10 to the minus third leaders. It's the same thing. According to that table in the first chapter, I think it's the first chapter of this textbook. Um, right now we're no leaders. Now we're going to use the density to get to Graham's ransom 0.7 zeros your grams for every one millimeter. Then we could get two moles. So I've calculated the weight of octane to be 114 0.23 grams per mole. So we're doing great canceling out right now. We're in bowls of octane. Very good. The leaders cancel out leaders, cancel out. Awesome. Um, so now we're going to use the conversion between the the coefficients of the balanced equation. Right? So if you balance his equation for you'll know that they're 80 two's to everyone. Uh, mole of octane. So we're gonna do eight moles of CO two for every one bowl of octane, and then grams of co two, and then we're done with this part of the problem. Right? So, uh, the molecular weight of CO two was 44.1 grams for everyone. Mole bulls Cancel out. Now we're in grams of co two. So this is the answer. The first part of the problem, which is how many grams can possibly form, right? So without rounding right away, um, doing this math, I was able to get, um for two thousands. Yeah, 21. Sorry. 2157 points. 54 grams. Right? Um, And when you ground this 23 significant figures which is needed for the answer, uh, you can report this is 2160 grams of co two. So here's my issue with these parted problems is that do you use the, um, the answer. So we have to report each answer to each part with the right number of significant figures that looks like that's three based on the numbers that are given. So we're gonna report our answer that the way you would report this on a homer problem if you have to account for significant figures is using this right. But the philosophy of answering these questions is that you take the answer before you don't want around that any point in the problem. Um, before you calculate before the last calculation and get your answer. So you have all of the you have the most accurate answer before you around, right? So if we really used 2160 as our theoretical yields, as opposed to 2157.54 um, your answer might be different, right? So I've done this. I've done both, and the answer is actually only off by 0.1% which could be big. Depending on your teacher. That could be all you need to lose credit. Right. So let me start a second notebook here, Right? So percent yield. Just to remind you, um, is the actual I'm writing like a monster theat actual products that you observe over the theoretical product that's possible to form based on the street geometry of the reaction. Right. So we were told that 19 Sorry at one point. Sorry. 1.9 times. 1.0 times. 10 to the third grams of co. Two were observed. That's the actual so aptly observed formation. And we calculated, um, 2157.54 grams of CO two. So we're calculating the percent year. We've gotta multiply this by 100. So the answer that I got was when you do when you use 21.57 you get 80 eight 0.1% yields. If you use, um, 21 sixties if you use your rounded answer from part, eh? Um, you you you you would get 88.0% yield. Um, So I, um I would suggest you can round your answer for reporting the answer, but then use the more accurate answer when calculating your yield is my opinion. So I would report the answer is 88.1. Um, if I were grading this, so yeah, that's the answer to this problem. Just, you know, significant figures are important, but make sure that you're reporting the right answer, right?