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Which of the following pairs of compounds can be used Ilustrate the law of multiple propontionst NH: and NHAC ZnOz and ZnClz HzO and HCI NO and NOz CHa and COzyour...

Question

Which of the following pairs of compounds can be used Ilustrate the law of multiple propontionst NH: and NHAC ZnOz and ZnClz HzO and HCI NO and NOz CHa and COzyour choice your Own words ~ Exploin0 grams oxyben, 10.0 grams of carbon,and 20.0 grams of chemical X is found to contain sample of chemical : should contain sample proportion would predict that a 70-gram nitrogen_ The law of definite grams = of carbon? how many 5.0 grams 7.0 Brams Brams 15 grams 20 gramswork _ Show yourto the Law of con

Which of the following pairs of compounds can be used Ilustrate the law of multiple propontionst NH: and NHAC ZnOz and ZnClz HzO and HCI NO and NOz CHa and COz your choice your Own words ~ Exploin 0 grams oxyben, 10.0 grams of carbon,and 20.0 grams of chemical X is found to contain sample of chemical : should contain sample proportion would predict that a 70-gram nitrogen_ The law of definite grams = of carbon? how many 5.0 grams 7.0 Brams Brams 15 grams 20 grams work _ Show your to the Law of conservation = possible . according - reactions are NOT of the following chemical Which = mass? 1. CCla CH: 3Hz 2NH; 2H,0 3. 2Hz Explain your choice: water Vapor from the air to form with oxygen = hydrogen = reacts of hydrogen gas combustion oxygen reacted? 4.In the water, how much water oxygen *10 g ofe Hydrogen and produce hydrogen burn 20.2 g of you



Answers

Write the balanced equation, then outline the steps necessary to determine the information requested in each of the following:
(a) The number of moles and the mass of Mg required to react with 5.00 $\mathrm{g}$ of HCl and produce $\mathrm{MgCl}_{2}$ and $\mathrm{H}_{2}$ .
(b) The number of moles and the mass of oxygen formed by the decomposition of 1.252 $\mathrm{g}$ of silver(I) oxide.
(c) The number of moles and the mass of magnesium carbonate, $\mathrm{MgCO}_{3},$ required to produce 283 $\mathrm{g}$ of carbon
dioxide. (MgO is the other product.)
(d) The number of moles and the mass of water formed by the combustion of 20.0 $\mathrm{kg}$ of acetylene, $\mathrm{C}_{2} \mathrm{H}_{2},$ in an
excess of oxygen.
(e) The number of moles and the mass of barium peroxide, $\mathrm{BaO}_{2}$ , need to produce 2.500 $\mathrm{kg}$ of barium oxide, BaO
$\left(\mathrm{O}_{2} \text { is the other product. }\right)$
(f)

Using the information provided. Let's determine the chemical formulas of each of the compounds and then draw their Lewis structure. So we have five compounds which are listed A, B, C, de and E. So let's first cell for our compound d. Um, because we're told that Compound D is 43.7% nitrogen and 50 percent oxygen. So for the mass of hydrogen, it will be 100% minus 50 percent minus 43.7% which would be equal to 6.3% hydrogen. So we're also told that the density of Compound D is equal to 2.86 grounds per leader at STP. So no, let's go ahead and find the formula for de. We can find the moles of n which is going to be, um, before we do that, let's assume that we have 100 g of D. And so we have 43.7 grams of N 50 g of oh, and 6.3 g of hydrogen. Let's convert each of these two moles, uh, 14 g per mole, smaller amounts of nitrogen. This would give us all right? Uh, yeah, 3.12 malls on 3 16 g per mole. So 55 about 16 gives me 3.12 malls and times 1 g per mole. That's equal to 6.3 moles we're going to divide by have 3.12 This would give me 11 and two. So my empirical formula for Compound B would be called to, uh and oh h two and we can solve for the empirical mouse for D, which would be equal to. And that's very each to their N O H. Two. So 32.0 g per mole. We're told that the density is 2.86 grounds for leader. So, uh, to let's hope for the molar mass of D. We have 2.86 g per liter times 22.4 liters per mole and this would work out to leaders would counsel 2.86 telestrate. You play for 64.1 pants per mole eso for n which is our whole number ratio. Mm. Over e m 64.1 grounds per mole, divided by 32 g per mole. And so we killed the two. So the molecular formula for D would be equal to two times the n o H. Two, which should be equal the end to O to h two. We rearrange this, we can get a compound which is equal the n h four n 02 So there is Compound de and H four to. Okay, so now there's d. We can identify compound. See, we're told compound. See, it has one more oxygen full, then compounds mhm d. So this is equal to N H four and 03 for the formula of C. Um, Now let's move on to compound A. We're told that compound C and a half one iron in common eso has one I am in common with compounds. See on when is acting is a strong electrolyte. The solution is strongly acidic. Eso it's strongly acidic. If it is strongly acidic, a strong acid would be nitric acid. The ironing common would be the nitrate iron. So there's compound a Andi, we've got Compound B, which we can do next. And here we have the tie. Trish, in of Compound B, requires 20. So whatever compound B is, um, it's gonna be tight treated with hydrochloric acid for complete neutralization. So I know that this has to be a base so tight, traded with HCL and obtain The Moller mouse s 07 to 6 g of compound B is tight, treated with 21.98 There's a neutralization, their middle leaders of mhm one Moeller hcl The moles of HCL is equal to one Moeller times 0.2198 leaders and this is equal to a point 0 to 1 98 moles of HCL and what we're going to assume here is we're going to assume a We're gonna assume a 1 to 1 more ratio, so we're going to assume 1 to 1 more ratio. So we have 0.2198 moles of compounds be and there for the Moeller mass of be would be equal 2.726 g over 0.2198 moles And this would work out to 30 1976 33.0 grounds per mole is the Moller massive beat a compound Be, you know must be basic since there's a neutralization. So it has to contain an O H group and ohh the O. H. Group is 17 g per mole, so 33 grounds per mol minus 17 g per mole will leave us with 16 grounds for more and so on a compound B. Uh, since it must be basic and it has to be 16 g from all the 16 g per mole must be comprised of N H two. Therefore, Compound B is an H two oh each and there's the identity of Compound B. And lastly, Compound E. It's as commercially available concentration solutions of air normally 16 Mueller Um, yeah. So, uh, compound A, we're told, is h N 03 and commercially available A 16 Moeller the solution of Compound E, which is 15 Moeller Um Hmm. Which company? E. Which is 15 Moeller I mean on this Must be but be solution with commercially available. Um, all the comrades are strong electrolytes. If it is a strong electrolyte 15 Mueller, This would be n h four Oh h Now let's draw our Lewis structures for compound A component E was H and 03 and it's Louis structure for nitric acid is and plus double bond. Oh, oh, minus Yeah, yeah. Oh, and H There's a Louis structure for a for B way have N h 20 h, which is hydraulic saw aiming on Lewis structure here O h and lone pairs here or compound. See, we have ammonium they treats, which would be ammonium, the ammonium iron and nitrates, which is in Yeah. Oh, this would be a minus there on the O double bond. Oh, lone pairs ceremony. Nitrates, compound D or Sorry, ammonium nitrate in each four. No end to their Oh, no. Sorry for and through. They get it wrong. NH 403 Let's make this correction here. This is N H four n 02 Let's fix. Come. Um, this should be nitrates, not nitrate. I hear So nitrate would be and plus couple bond. Oh, lone pairs. Oh, whole minus minus And ammonium nitrate B n plus H h h h. And things should be and plus oh, minus on pairs Double bond. Oh, so ammonium nitrate and compound E waas ammonium hydroxide So and plus and a drug side Oh h minus sell ammonium hydroxide there. So those are the five compounds and their Lewis structures. Yeah,

In order to find our unknown compound, we're going to have to use a little bit of math to understand the relationship. Between are various elements and we're going to be using Avery Simple systems of equation to understand how we would find the moles of our compound. So let's first note that our compound X is containing carbon, hydrogen and oxygen, and we're told we have this reaction where we have X Plus three die atomic oxygen that's going to be forming two moles of carbon dioxide plus three moles of water. Now, this is where it gets a little bit confusing mathematically, um, so we're going to have to find the number of atoms of each carbon, hydrogen and oxygen, and this is the reaction we're going to use To do that. We know that c sub x h sub y o sub z plus X plus y over four might have see over to forms X times carbon dioxide, plus why, over two times water. This seemed a little bit complicated, but all these variables are saying is that we have coefficients in front of our products and reactive, and we have a number that denotes number of atoms in our compound. From this equation, we can deduce that X equals two. And why, equal six? This was found by simply comparing this equation to the one above. And now, when you defined Z so we'll get exposed. Why Over four minus Z over to is equivalent to three. And what this means is, when we simplify, it is he is equivalent toe one. So that means our formula is C two h 60 Now, the way we know that this is a molecular formula, not an empirical formula, because remember, we're dealing with di atomic oxygen and because oxygen is die atomic, we could have various empirical formulas. So this is how we found the number of atoms in our formula. Now our job is to draw our electron dot structures of how this compound can form. This first one will have a CH three group that's bound to an oxygen that's bound to another CH three group. And our second structure we can possibly have is a ch the three group bound to a CH two group and that then bound to a hydroxide group using the same formula. But we just rearrange them differently now. The next step is we need to introduce ourselves to a concept called the entropy of Formation and the IMF appear formation is simply heat that we use or put into our reaction to former products. Remember, this is our new reaction because we found her unknown compound. We have C two h 60 plus three die atomic oxygen that's forming two moles of carbon dioxide and three moles of water the way we would find our en Fille Pierre formation, which we denote with this symbol Delta H not of F. We'll take our products minus the reactant since so we'll have to times the entropy of carbon dioxide, plus three times the entropy of water minus the infill P of our compound C two h +60 Now, if you're confused as to why we don't have 302 in our formation, it's because it pardon me, it's in it's elemental form, so we're not going to be including it now. We just simply plug in the values of entropy, a formation for each compound, and we'll have that. The entropy of formation is equivalent to two times negative. 393.5 plus three times negative 241.82 plus 1326.6. Now it's just a very simple calculation and we'll find that the entropy of formation of our compound is negative. Negative. 185.86 Killer jewels promote. So I hope with this problem problem helped you understand how we can find an unknown compound by simply knowing the atoms that make up that compound and the overall reaction. And I hope of this introduced you to how we can rearrange dot structures and hopefully you now understand how we confined the entropy of formation, which is a very important concept.

Let's start this problem by remembering what the ideal gas law is. We sometimes call this the pivot equation. This is saying PV is equivalent and R T P is our pressure. V is, er volume. And as her number of moles are simply a gas constant and T is her temperature, we want to know the moles in this problem. So we're going to rearrange this equation to end by itself. And when we do that, we'll get N is equivalent to P V over R T. So the next step we want to know is the most of Arm Ethel effort group, and this is the way we're going to do it. We're going to have to do a very small conversion in our calculation, will get that end is equivalent to 100 times 10 to the third multiplied by one over 101,325. 8 teams will deployed by one will divide that by 10.8 to 1 times 327.8. All that is different in this calculation than maybe a different calculation with Prevnar is we have to you change our pressure units into a t m's. And when we do that and I do, a calculation will get that The moles of Arm Ethel Effort group is 0.366 moles. Now we know that one mole is going to have 88.33 g. This is simply by comparing it to the mass of our sample were given. And if we want to know one mole of Arm Ethel, other group, What can we produce from this Will take one. Pardon me. 0.15 divided by 88.33 will get 1.7 times 10 to the third power. And now what we can do is we confined the moles of our reaction essentially by saying five multiplied by 8.5 times 10 to the negative third, divided by 1.7 times 10 to the negative. Third, this part of our expression represents our moles of carbon dioxide. Well, multiply that by 10 times 10 to the negative Third, divided by 1.7 times 10 to the negative Third. This part of our problem represents arm Ethel, eh, thermals. When we do this will find that we will produce six moles of water in her reaction. So by comparing the moles of each of our products or reacted in our reaction, we confined our empirical formula, which will be C five h 12 0. Now we want to know well, what's going to be our molecular formula. Is it going to be the same? Are we going to have a different number of Adam's? It's going to be the exact same as our empirical formula, because our Mueller weight is roughly 88 which is the same masses we had before. Now on to Part C, we're told. Let's write their reaction for our compound. And remember, we already calculated six bowls of water, and we already knew that we had five moles of carbon dioxide. So this is should be just filling in a quick blink. We have C five h 12 0, which is the compound we found before. Well at add eight moles of diatonic oxygen will produce five moles of carbon dioxide and six miles of water. And now, for Part D were told. Let's find the entropy of formation of this reaction, and the entropy of formation is just simply saying how much energy or how much heat is our reaction need to create our products of carbon dioxide and water. And if this is a new concept for you, we didn't know Delta H as our entropy of formation are a little look Looks like a zero Here is actually what we call not to. This is saying that we're at standard state conditions and the equation for entropy is equivalent to the sum of the entropy of our products. Minus is some of the entropy of our reactive and the entropy for our reactions or products are known values. All we have to do is plug them into our equation. So we'll get that the entropy of formation will be five times negative, 393.5 plus six times 285.8. We'll subtract a negative 3368.7. And then this is just a simple calculation to find that are in Philippi of formation for our reaction. And the formula we found is negative. 313.6 killer jewels per mole. So I hope with this helped you understand how we can use the concepts of gashes chemistry and apply it to concepts like combustion. And hopefully this problem allows you to see how we can find the empirical in molecular formula, knowing mass and moles. And hopefully this introduced you to a concept called the entropy of formation.

But this problem were given a balanced equation of a plus five B yields three C plus four D Um, and we're told that equal masses so the mass of a equals the mass of B R reacted, and we're trying to figure out which will be the limiting reactor based on two different scenarios. In the first scenario, um, it says that a if the Mueller massive A is greater than the molar mass of be, can we determine what the limiting reactant is? And the answer is we cannot determine limiting. Oops, reactant, because even if a is greater than be, we cannot be sure that a oops that a will be greater than five times be so because we don't know that it will be greater than five moles of be we cannot be sure if a will be a limiting reactant or fee will be the limiting reactive. However, in the second scenario, where it says that, um, be the molar mass of B is greater than the molar mass of a. Then we can be sure that B is the limiting reactant the dent because, um, if the molar mass of be is greater than a then five b is greater than a so therefore, that means that there will be fewer moles of be so it will be The limiting reacted reacted in this scenario. For the next part of the question again, we have our equation of a plus five b produce this the re C plus four D. We're told that C is c +02 and D is h +20 so we can rewrite this equation as a plus five b produce this three CEO too plus for H +20 which is a combustion reaction. Um, but where they're asking us if a has a similar Muller master carbon dioxide and B is a diatonic molecule. We want to identify be on DSA Porter. Answer now, since it's a combustion reaction to the dye atomic compound in combustion reactions is oh too. But let's go ahead and prove that 02 is actually gonna be what this is stuff. First thing we want to do is find the molar mass of CO two. So CEO, too, has a molar mass of 44.1 grams per mole and H 20 has a Moeller massive 18.16 grams per mole so we can find that on our react or on our products side, we're gonna have three times 44.1 plus four times 18.16 which gives us a total mass of 204 points. 094 Um, And with based upon the law of conservation of matter, if there are 204.94 grams on the reacted or on the product side, that's how many have to be on the react inside as well on and also said that a is similar to carbon dioxide so we can take this 204 0.94 and subtract if we just subtract 44 because you say it's similar. That's gonna give us 160.94 for the mass left by B. If we divide that number by five, we get 32.188 Um and we're again. We're told it's a dye atomic molecules. So divide that by two and we get about 16. So be is going to be, oh too. So we know that for sure. Next part of this problem. Now we can go ahead and put in even more information into this equation. We have a plus. Um, 502 produced this three CEO too. Plus four h +20 And we're trying to find the empirical, um, and molecular formula for a So And we're also told also given the information that A is a hydrocarbon that has 81.71% Harbin by mass. So we know that A is about 44 because it's similar does Theo too. So if we multiply that by 0.8171 we get 35.95 And if we divide that by the molar mass of carbon 12.1, we get 2.99 So approximately three. So we know it's going to be C three, but we still don't know how many hydrogen are gonna be in it. So what we can do here now? Since we know we're gonna have three carbons, we can take 44 and subtract three times 12.1 which will give us about 7.97 and we know it's gonna be hydrogen. So if you divide that by 1.8 we get approximately 7.91 which is pretty close to eight. So our formula is going to be C three h eight, and since it can't be simplified any further, it's gonna be both our empirical and our molecular formula is C three h eight.


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