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14) The theoretical yield of SigN4 (in grams) if 1.50 Moles of Si and 1.25 Moles of Nz in the Reaction:3 Si + 2 Nz SizN4a) 0.5 mole 87.5 gb) 0.625 molec) 70 g...

Question

14) The theoretical yield of SigN4 (in grams) if 1.50 Moles of Si and 1.25 Moles of Nz in the Reaction:3 Si + 2 Nz SizN4a) 0.5 mole 87.5 gb) 0.625 molec) 70 g

14) The theoretical yield of SigN4 (in grams) if 1.50 Moles of Si and 1.25 Moles of Nz in the Reaction: 3 Si + 2 Nz SizN4 a) 0.5 mole 87.5 g b) 0.625 mole c) 70 g



Answers

For the reaction shown, calculate the theoretical yield of product in moles for each of the initial quantities of reactants. $$2 \mathrm{Mn}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MnO}_{3}(s)$$ (a) $2 \mathrm{mol} \mathrm{Mn} ; 2 \mathrm{mol} \mathrm{O}_{2}$ (b) $4.8 \mathrm{mol} \mathrm{Mn} ; 8.5 \mathrm{mol} \mathrm{O}_{2}$ (c) $0.114 \mathrm{mol} \mathrm{Mn} ; 0.161 \mathrm{mol} \mathrm{O}_{2}$ (d) $27.5 \mathrm{mol} \mathrm{Mn} ; 43.8 \mathrm{mol} \mathrm{O}_{2}$

In this question, we're going to be looking at stake your metric particular interesting more two more questions. Okay, so in terms of theoretical yields and limiting ratings. Say we've got a that is reacting with B from C N. So what is happening in terms of limiting reactant? The limiting reactant is that one reacted in which the reaction depends on. So say we've got a that is a lot. We've got more of A than be. So in this case B is going to be the limiting reactions. We are saying at some point we're going to have a situation where by all of the B is going to be consumed and they're still going to be some A that is going to be left. That is A is in excess. So what then happens is this creation is going to stop even if we still have a lot of A that is left because there is no more be that is being that is available for it to react with, you know, God. So we're saying this reaction that is the production of C entity. It depends all the limiting react, which in this case is big because it is going to be consumed first and then spoke the reaction. So whatever great career field of the product it is going to end on limiting reactions. So what we're going to do here is to first of all look at the limiting reactant and then conclude that is going to give us our theoretical you So say we've been given three Mm Plus or two um And or three am entering or Yeah. So what we have here, we look at and then Against N three or 4 We can tell that we are three sure one and if we're looking or two yes. And we can tell that we are in two is 2, one in terms of more. So what we do here is to look at the amount of Mm in three or 4 that we're going to form through if M N was the limiting reactant and if an auto the limiting vehicle. So I'm playing this we are saying say we've got three moles, You've got three malls the moors off mm saying the number of more off M N three or four is going to be 1/3 of the number of moles of and and this is going to be one. Yeah, but if we are to look at this in terms of or two, you're saying the number of moles of and in three or for is equal to 1/2. 1/2 multiplied by the number of malls of or true number of wars. Yeah. And this is going to be 1/2, multiplied by three more And this is going to give us one point. Hi, so from here we can tell that we are forming a lot more of em and or three if we are Looking or focusing on or two in this case we have or two that isn't exit or two is Yeah. Access and mm And is the limiting create therefore if we are to look at this from the limiting reactant perspective, the theoretical yield is going to be the theoretical Yeah Oh M N three or 4 is going to be one. No, So more than one. If we are to look at Mn we can tell that the number of walls of MN three or 4 is equal to one of the three multiplied by four. And this is 1.33. And if we have to look at this from all to perspective, the number of moles of them in three or four Is going to be equal to one of the two, multiplied by seven. So this is going to be equal to three right? Yeah. So at the end of the day, we can still tell that mm Yes, limiting create tent therefore deal M N three or four, It's going to be equal to one 33. Yeah, because the union is determined mostly or it is driven by limiting reactant. So moving on to the final one. If we look at this from the perspective and we're going to form the number of morals or MN three or 4 going to be equal to one of the three divided, multiplied by 7.5. NBC's 9.17. What? And if we look at autumn number of moles of mm three or four is equal to half seven and this is equal to one name. Yeah. Again, And he's the limiting reactor. Therefore, the yield, the theoretical yield of MN three or 4 is going to be equal to nine one second. Yeah.

In this exercise, we are asked to find theoretical yield. These are all pretty simple calculations. Were given the equation, you see right here, three manganese React with two oxygen's to produce one manganese for oxide. For our first scenario, we are given three bowls of manganese And we are given three moles of 02. So we have to determining our determine are limiting reactant. So I'm going to do two quick calculations 1st. I'm gonna start with three moles of manganese. I'm used the mole ratio from the balanced chemical equation three moles of event in one mall of MN 04 To get 1.0. I'm just gonna go 1.0 malls of MN 04. I'm going to do the same calculation with moles of oxygen. Of course, I'll have to use the mole ratio, which will be one more of product and two moles of oxygen, Giving me 1.5 moles of product. So this is how much I can produce. Remember. The smaller of the two values in this case will be our limiting reactant. And that limits what we can produce. For the second scenario. We are given four moles of manganese And we are given seven moles of 02 So, again, we're going to set up the exact same equations, and since I'm setting the exact same equations up, I'm going to leave the substances off. That's hard for me to do. I'm going to put the M in in. I can't do it, sorry. And this gives me 1.3 moles of product On this side. I'm gonna start with seven moles of 02. Again, I have a 1-2 ratio of product to react it, and that equals 3.5 moles of product. Again, we pick the smaller of these to determine what the theoretical yield is. And her last scenario is 27 .5 moles of agonies and 43.8 moles of oxygen. Again, we set up, whoopsie not the exact same. We set up the same ratio 27.5 moles of manganese, And it will be 1-3 ratio. This one equals 9.17 malls of product, And we have 43.8 balls of oxygen. I'm going to multiply by that by the 1-2 ratio of product two reactant In this one equals 21.9 malls of product. And for the third time we pick the smaller of the two numbers. So here's a theoretical yield for each of these.

So for these problems, we need to calculate the molar mass for a substance and divide the molar mass from the ground to get the moles part. Part part A. We have 50 grams of organ car gone. Whatever growing up in a more mass, which is, um, let's see 39.5. Sorry. 39.95 grams. This is the Moeller maps. This is equal to second. Can't see the type Booker 1.2 five moles off are gone. Part B 0.2 grams of carbon. Two girls divide that by the molar mass of 12.1 grams per mole. The Marlins Muller Mass Do I have that? And we get the molds of Karpin, Part C 15 grams of and h three. So any three is 14.1 close three times one point no one for the hydrogen, 17.4 grams. The more mass to find that secrets of 0.8 moles of NH three and party It is 75 grams of us too. So ass is 32.6 plus two times 16 for the oxygen's in the 64.6 grams. This is a more mass equal to 1.17 miles. Well s 02


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