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Consider Ihe reaction _2Pb(s) + 0,(g)2 PbO(s)Eof lead, forming 369.9 g of lcad(Il) oxide. Calculate the pcrcent , yield of An cxcess oxygcn reacts with 4514 the rca...

Question

Consider Ihe reaction _2Pb(s) + 0,(g)2 PbO(s)Eof lead, forming 369.9 g of lcad(Il) oxide. Calculate the pcrcent , yield of An cxcess oxygcn reacts with 4514 the rcactionpetcent yicld:

Consider Ihe reaction _ 2Pb(s) + 0,(g) 2 PbO(s) Eof lead, forming 369.9 g of lcad(Il) oxide. Calculate the pcrcent , yield of An cxcess oxygcn reacts with 4514 the rcaction petcent yicld:



Answers

Lead(II) oxide is obtained by roasting galena, lead(II) sulfide, in air. The unbalanced equation is:
\begin{equation}
\mathrm{PbS}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{PbO}(\mathrm{s})+\mathrm{SO}_{2}(\mathrm{g})
\end{equation}
\begin{equation}
\begin{array}{l}{\text { a. Balance the equation, and determine the theoretical }} \\ {\text { yield of } \mathrm{PbO} \text { if } 200.0 \mathrm{g} \text { of } \mathrm{PbS} \text { is heated. }} \\ {\text { b. What is the percent yield if } 170.0 \mathrm{g} \text { of } \mathrm{PbO} \text { is obtained? }}\end{array}
\end{equation}

So to sell this question, the first thing we need to do is balanced this unbalanced equation and to balance any equation. It's always good to start with Oxygen's and we compare both sides. So on the left side we see two oxygen's and on the right side we see three. So you always look at the the compound that has an odd number of oxygen's and you make it even and we do that. Bye. It's and we do that by increasing this amount, the amount of this compound by two and then So we added to lead oxides. But now we see that there are two leads on this side on the right side, but only one lead on the left. So we go ahead and we right to on the left side for lead sulfide. But now we see that there are two silvers. So we go over to the right, we see s 02 and we right to here as well. So lead and sulfur or now even. But now we change the number of oxygen's we multiplied s 02 by two. And there are two oxygen's in, uh, eso to. But now there are four. So we look at lead oxide. We see there to oxygen's here now, so we add them together for plus two is six. There are six oxygen's on the left on the right side. So on the left side oxygen there are only two here. So we multiply that by three, because three times two or six So now this equation is balanced. Oh, that's first bit First part of a now the second part of a second part of a we need to find the theoretical yield of P B O ah, blood oxide. If there are 200 g of let's sulfide being heated, I just got used. So in order to find the theoretical yield, the first thing we need to do is convert this amount into moles. So we do that through a dimensional analysis. So we have 200 g PBS Multiply this by the Mueller Mass. So one mole, PBS is 239 grams per mole. So then we cross out like units and you calculate that and we get about 0.8 about 85 7. But PBS now, in order to find the amount of P B o We have to multiply this by the ratio off PBS to P B. O. So we take a lot of the ratio. There are two PBS, uh, to that sulfides for every to lead oxides. So it takes two molecules off lead sulfide to create two molecules of lead oxide. Conveniently. Since they're the same, it's just a wonderful one ratio. So Okay, if we do this in full, we have 0.8357 moles of PBS. Multiply it by the ratio cross outlet units again of the and then we get the same number. Okay. And there we go. 0.8357 moles of P B O. And then you would use the mental analysis again to convert the mold into grams. And the molar mass of P. D. O of lead oxide is 223 g. Come. And then we calculate that Yeah, we get 1 86 0.5 for Gramps. Uh huh of lead oxide. But then you would divide this number by the mass off sulfur up lead sulfide, and we were given 200 g, right? So you would divide lead oxide. You divide this mass by the mass of of PDS led sulfides as 200 and we get our theoretical yield, which is zero point 93 But then you convert this into percentage, it becomes 93%. Right? Okay, Okay. Help! So there are answers for part A and part B. We want to find the percent yield. If we got 1 78 g instead. What do cool. So all we need to do is just divide this by that mass. We don't have to do anything else since the ratios of the same. Okay, divide these. And you would get about 0.85. Although I'm doing this wrong. No, I'm not doing this wrong. This is right. So this is 0.85. This is about 85%. It's or not. This is 85%. So in order to get the percent yield, Okay, Yeah. You have to divide the actual yield by the theoretical you'll and multiply that by 100. So the actual yield is 85%. We were just right, 85. It's theoretical. It's 93 and we get okay, okay. 91 but then multiplied by 100 and we get 91%. The full decimal will be 91.3%. Because this is 0.913 Yes, those are your answers.

This problem is also a limiting reactant problem where we are given to amounts for two different reactions and we need to figure out which one is the limiting reactant and then from the limiting reactant to determine the theoretical yield and ultimately the percent yield. The balanced chemical reaction is two moles of iron. Three oxide reacts with three moles of carbon to produce four moles iron And three most carbon dioxide. So if one of the reactant is iron, three oxide and we have 12.5 g of it, we can convert those grams into molds by divided by the molar mass of iron three oxide than knowing the moles of Iron three oxide required to produce four moles of iron. This being the ST geometry, we can calculate the molds of iron produced, then multiply by the molar mass to convert the moles iron into grams iron and we get 8.74 g iron produced from 12.5 g iron three oxide. Next, we'll do the same thing with carbon. We have 2.6 g of carbon, which we can convert to moles carbon by dividing by the molar mass carbon. Then, when we know the moles carbon, we can calculate the moles of iron With the 3- four strike geometric relationship. Then when we have moles, iron will multiply by the molar mass iron to get graham's iron And if all 2, 6 g of carbon were consumed would get 16 g of iron. Therefore, iron three oxide produces the least amount of iron. This being the theoretical yield 8.74 g and iron three oxide producing the least amount of iron is the limiting reactant Percent yield will be what we actually get, which is provided in the problem, divided by the theoretical multiplied by 100 and we get 80.1 yield.

Custom 61. Yeah, the reaction do encourage him outside on carbon dioxide is given C E T or three 14.4 g of Branch Motel on 13.8 g of family else here or I love to react. And once reaction was completed. 19.4 g off the Carson carnage Waas Collective The limiting reacted theoretical year on a person is years off. The reaction are to be estimated, so that is, look at the limiting reaction first by comparing the moral muscle. Beach 56.1 is a modern months off can see multi. 44 for somebody else on 100. Indians 100.1 Garcia carnage. So that means 56 plan for 1 g will react completely with about 4 g toe produce 100 point 1 g. So let's see how much off governor activities required for 14 point for them. So 41 4 g needs 14.4 multiplied by for people upon 56.1. Yeah, 14.4 multiplied by 44. By practice, it's 440.1 g. Aw, really? Germany Option? Yeah, that we don't have to be 11.29 Jin's actual amount of carbon else. It is 13 point babies. So some members of company also be reading Object. That's the market is completely converter. That means God's, um outside is the limiting Lee. Is it now from here, difficultly, they well can play the killed Agreed. So 14 point 46.1 g will you? 100 year 100.1 g of plants and gardening. So 14.4 with you. 14.4 What he played 100. Find one. Ah, for 56.1. This equals 25.69 graham toe Guys in memory. The actually will obtain this 19.4, you know. So the person is your liquids. Actually, upon the article in 200 he quit 19.4 a whole 25 onesies equals 75 point by two 75 born height. Listen

This is a limiting reactant problem where you're given amounts for two different reactant and you need to figure out the mass that each reactant can produce a product and which everyone produces. The least amount of product. That's the limiting reactant and the amount of product that the limiting reactant creates is the theoretical yield. So let's start with the 14.4 g of calcium oxide will convert to moles calcium oxide by divided by the molar mass. Then used to study geometry to go from moles calcium oxide. Two moles calcium carbonate, then multiplied by the molar mass calcium carbonate. To get graham's calcium carbonate, we can do the same thing with the carbon dioxide that we have will divide by the molar mass carbon dioxide to get moles carbon dioxide. Use the stock geometry to go from moles carbon dioxide two moles calcium carbonate, 1: one. Then we'll multiply by the molar mass calcium carbonate to get Graham's calcium carbonate. So the calcium oxide produces the least amount of calcium carbonate. Therefore, it is the limiting reactant, and the amount of product that it creates is the theoretical yield. The % yield will then be what we actually got in the laboratory, provided at 19.4 in the problem, Divided by the theoretical yield times 100 And we get 75 5 yield.


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