 Okay. Let's try this one. So, similar to the last one that we did, what is the maximum mass in grams of N-O that could be obtained from 15.5 grams of dinitrogen tetroxide and 4.68 grams of dinitrogen tetrahydride when they react. The balance chemical equation is... okay, so let's go ahead and do this. So, let's write down what they give us. So, it says the mass of N-204 is 15.5. That also gives us the mass of N-2H4 4.68 grams. Maximum amount. So, the thing we're going to have to do, again, is figure out the moles and moles and then figure out which one's the limiting reagent. And then we're going to have to go to the other side of the equation. Okay, so, just like last time, we're going to have to convert it to moles. So, grams per mole. Okay, so for the first one, I got 92.02. Did I get that one right? Yeah. Okay. So, let's write it right. Okay, they're right. Okay, so let's do the other one. So, 14.01 times 2 plus... Okay, so let's figure out the number of moles. We've got each of these. So, 15.5 divided by 92.02. 1, 6, 8 moles. Right, so this one we need to do that last bit of converting. So, we could change whichever one we wanted, but let's just do the top one. So, we've got 2 moles, N-204, every 1 mole. So, for this many moles of N-204, we're going to need to divide this number by 2. So, add that much N-204. Yes, right? We have this much. So, that's more than this. Is everybody cool with that? Yes. So, this one here must be the limiting reagent, right? Yeah, you can keep going with it. Yeah, definitely. And in fact, pardon. Yeah, and then you can do the mass after that. We're just doing it step-wise, step-wise. You know, this is kind of basic step-step-step-step. But yeah, it's nicer, especially if you've got a real long board, you know, you can go all the way, you know, step-step-step. Okay, so limiting reagent is this. So, this is the number of moles we're going to be working with here. Okay, so, the question asked to us, what's the maximum mass of N-O? That can be made. So, if we've got the number of moles, figure out the number of moles of N-O. And then, the maximum amount in grams, well then, we'll just take one mole and what's the grams of N-O of N-O? It's going to be 14.01 plus 16. So, effectively, hopefully you see it's the same thing. You just stuck this part onto them here, right? So, when we do that, we get times 6 divided by 2 times 30.01. The maximum mass in grams that you could get from this is 15.2 grams of nitrogen monoxide. Do you guys get that one? Same answer, very close, at least. Any questions on that?