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Ammonium nitrate, NHANOz undergoes decomposition on heating to give nitrous oxide; NzO gas and water: Write a balanced equation for the reaction: In a certain Exper...

Question

Ammonium nitrate, NHANOz undergoes decomposition on heating to give nitrous oxide; NzO gas and water: Write a balanced equation for the reaction: In a certain Experiment; student obtains 0.340 L of NzO gas at 718 mmHg pressure and 24 %C. calculate the value of the gas constant; R:Write the extended electronic structures [showing spins of the electrons] of the following; Fanion. the atomic number of F is 9 S atom; it has [6 protons in its nucleus Potassium; K; its atomic number is 19.Show the str

Ammonium nitrate, NHANOz undergoes decomposition on heating to give nitrous oxide; NzO gas and water: Write a balanced equation for the reaction: In a certain Experiment; student obtains 0.340 L of NzO gas at 718 mmHg pressure and 24 %C. calculate the value of the gas constant; R: Write the extended electronic structures [showing spins of the electrons] of the following; Fanion. the atomic number of F is 9 S atom; it has [6 protons in its nucleus Potassium; K; its atomic number is 19. Show the structures of methane CH:; formaldehyde, CHz0 and acetylene, CzHz exhibiting the formation of &-Bonds and T-Bonds, type of hybridization and overlapping of orbitals. How many O-Bonds and T-Bonds are there in each of the three molecules?



Answers

Nitrous oxide $\left(\mathrm{N}_{2} \mathrm{O}\right)$ can be obtained by the thermal decomposition of ammonium nitrate $\left(\mathrm{NH}_{4} \mathrm{NO}_{3}\right) .$ (a) Write a balanced equation for the reaction. (b) In a certain experiment, a student obtains $0.340 \mathrm{L}$ of the gas at $718 \mathrm{mmHg}$ and $24^{\circ} \mathrm{C}$. If the gas weighs $0.580 \mathrm{g},$ calculate the value of the gas constant.

For part A. Without a calculation. Let's predict whether the volume of gas is collected can be used to distinguish between the two reaction pathways. Um, this reaction, one reaction pathways any four and three, which would be composed to end 20 and t rich, too. Oh, so in this reaction pathway, one more of and 20 is formed from one more, uh, image for in all three. And in the second pathway, he had any for and all three, which is going to decompose to end too 1/2 to and to H 20 So in this pathway, um, one more of into and 1/2 more of 02 um are formed from one more of N h 403 So get if in a church four n 03 decomposes to give a mixture of and to and 02 gases, then the volume of the collected gases will be more, says the volume of the collected gases if the reaction proceeds to create and to go so that a calculation. There's our explanation for Part B. Uh, let's we're told that the, uh, NH four displaces 40.256 grams just with 79 millimeters of water. Uh, is it into a or a mixture of and to go to? So which pathway would have beat? So, um, let's start with the 0.256 grams of N H. Four and three. Convert this two moles molar masses, 80 grams per mole. This would give 0.32 moles of N H four n 03 Now let's assume, uh, that it's going to be, too. Oh, still 0.32 moles of N H four n 03 on and it's a 1 to 1 more ratio. So one more and four and three as equipment, One more bend to host the sequel to you 0.32 Moles of End to Oh and, um, we can bring in Our STB conversion went 0032 malls of in 20 one, more equivalent to 22.4 leaders. This is equal to 0.71 60 leaders. And no, let's, uh, bring in Charles's law and go be won t one is equal to be to over t two. Plug in the numbers here 0.7 ones. See leaders. Temperature is to 73. Calvin V. 2 2020 degrees Celsius to 93. Calvin starting for the volume. Here we find the void. Here is 77 milliliters based on and 20 so let's look at the mixture of end to end 02 So if this is so, we have 0.32 Moles of the end for end 03 times. One mole of any four and 03 is 32 roles in the mixture, 3/2 moles of mixtures that's equal 2.4 eights. Moles of a mixture of enfuego no to and so 10048 Moles of the mixture. One more 22.4 liters of gas. That's going to be. Quote 2.1075 leaders again. Let's bring in Charles's law. Uh, 1 to 1. Be to t two 1075 Leaders 70. Propel. Vinh B 2/2. 93 12 in some for being too. Here we get 115.3 milliliters on this is beast on the mixture, which is N 20 to weaken. See here. Um, the question tells us 79 milliliters of gas is displaced. Therefore, the gas must b n 20 going back. Over here, 77 milliliters is closest to the 79 villages of gas displaced.

This question is a review of many concepts introduced in this chapter and also requires you to recall some concepts in previous chapters. It first wants you to consider the decomposition of solid ammonium nitrate to form gaseous, die nitrogen oxide and water vapor. And it wants you to balance this reaction, sell ammonium nitrate, ammonium with a plus one charge nitrate with the one minus charge requires one of each of them. Then we have the production of die nitrogen oxide die meaning to. So we have to nitrogen and one oxygen and water vapor. We have 12 nitrogen to nitrogen. So it's balanced, We have four hydrogen is, we have to hydrogen so we need to put it to their that gives us two plus one more three oxygen's three oxygen's. So now it's balanced. Part B asks you to determine the partial pressure of nitrogen or die nitrogen oxide and the partial pressure of water vapor produced in a 1.75 liter flask at 2 30 degrees Celsius. Well, to calculate the partial pressure associated with each of these at a known volume and temperature, we need to know the moles of each of these gases produced. We can calculate the molds produced by knowing the grams of ammonium nitrate that decomposed. The master decomposed is 2.50 g ammonium nitrate, which can be converted into molds ammonium nitrate by dividing by the molar masses of ammonium nitrate to mel's nitrogen plus the massive formals oxygen plus the massive three moles. Sorry, formals hydrogen plus the massive three moles. Oxygen Gives us 80.045g. Then, when we are in units of moles of ammonium nitrate, we can go two moles of either gas. We want to solve for first, I'm going to solve for the moles of die nitrogen oxide And we get 3.12 times 10 to the -2 moles of die nitrogen oxide, high nitrogen monoxide. It should be the mono on there, even though it's not found in the problem. Okay then we'll take the 2.50 grants of ammonium nitrate and with it decomposing, we can calculate the moles of water vapor that are produced. The first conversion factor is exactly the same. To get a smalls ammonium nitrate. Then we'll go from moles ammonium nitrate, two moles of water, recognizing it's not 1-1 is The previous problem, but 1- two. So we'll get twice as many moles of water vapor, 6.24 times 10 to the negative two moles of water vapor. Now, knowing the moles of each of them, the temperature and the volume of each of them. We can calculate the pressure of each of them, rearranging the ideal gas law pressure will be equal to end, which we just calculated are which is a constant, Multiplied by T. T. needs to be in Kelvin. So for at 230°C blood on to 73 to get it into Kelvin. And then we divide that by the volume P equals N. RT over V. And the V was provided at 1.75 leaders. So we get .7367 atmospheres as the partial pressure of die nitrogen monoxide. Then for water vapor, it's the same calculation but we're going to use this as our moral value Instead of the .1, 2 times 10 to the negative. To will use the point I'm sorry. We'll use instead of the 3.12 times 10 the negative two. We use the 6 to 4 times 10 to the negative too. That will double because that this is double the moles. Then this will give us double the pressure. The partial pressure of water vapor will be 1473 atmospheres part C. Then simply wants you to calculate the total pressure. The total pressure will be the sum of the two partial pressures, according to Dalton's law, partial pressure. Some of these two partial pressures gives us 2.2.210 atmospheres. And then, last of all, reviewing something introduced in the previous chapter, you're asked to draw three resonant structures for di nitrogen monoxide. To draw the resonant structures, you first need to know how many valence electrons are available per molecule of die nitrogen monoxide, Nitrogen has five valence electrons. There, two of them that gives us 10 Plus oxygen, which has six valence electrons gives us 16 valence electrons. It's also helpful to note, although not mentioned specifically, that oxygen is not the central atom. We actually have two nitrogen atoms bonded to each other and then the oxygen bonded to one of the nitrogen atoms. So if we were to bond these two nitrogen together with an oxygen bonded to one of them, I have chosen to do double bonds here first, using up to 468, recognizing I'm going to run out of electrons and we need to have some sharing occurring. So if I've used up eight, I have eight left over this, nitrogen has a sufficient number of valence electrons. This oxygen does not. It needs four more. This nitrogen does not. It needs four more so that the four that each of them need multiplied by two is the eight that I have remaining. This is one plausible resonant structure. Another plausible resonant structure might have a triple bond between the two nitrogen and a single bond for one of the oxygen's. This then requires the use of 2468 valence electrons have eight left over, so I can go 1234 5 6. Now oxygen has an octet. This central nitrogen has an octet. The peripheral nitrogen only has six, so it can use the two more that I have left over. And then the final resonant structure will be the triple bond between a nitrogen and oxygen and just a single bond between the two nitrogen. Again, I've used up eight for the bonding. I have eight left over. I'll put two on oxygen six on nitrogen. So everything has an octet and this is the last of the three um possible resident structures for di nitrogen monoxide.

All right for this problem. We're going to be covering balancing equations, writing ionic equations, types of reactions, finding maximum mass and percent yield this is the equation that were given. So the first thing we're going to want to do is balance this equation. And the way I like to do that is first I'm going to look for polly atomic ions that appear on both sides of the equation and I C N 03 on both sides of the equation. What that means is that means we can treat those as one component and not break it up and have to do all kinds of other math for that. So we're going to put the elements and components that we have over here first. So in we have a church And then we have our n. 0. 3 component. Alright, next thing I'm gonna do is I'm going to add all the components of these elements that we have. So we have one nitrogen on this side, four hydrogen On this side and one n. 0. three. Same thing on the other side. We're going to add it up one or and one. So that's already balanced. The next thing we're gonna do is we're gonna find the ionic equation which basically means we're going to take all of these components. We're going to break them up into their individual charges, find out which ones appear on both sides so we can cancel it and then look at what the reaction is really doing. So we can't break up this poly atomic ion here. So we're gonna put and age three which has a neutral charge. We can break up this into its hydrogen and N. 03 components. So we're going to have Plus hydrogen which has a positive one charge Plus n. 0. 3, which has a negative one charge. All right. And then that's going to yield we can break this up too at age four, which has a positive one charge. And we can add N03 With its -1 church. All right. So, that's our that's what we call our complete ionic equation. We're going to also look at the net ionic equation, which is what happens after we remove our Spectators or things that appear on both sides of the equation and therefore cancel each other out. So, the only thing I see here that will cancel out is N 03 since that appears on both sides. So, our net ionic reaction is going to be, sorry, equation not reaction is going to be NH three plus H Yields and H four. So, in essence, this is what this reaction boils down to. Alright, the next thing we're gonna do is we're going to figure out what type of reaction this is. Now a big hint is that this um this molecule here is called nitric acid. So, really that's an acid. So that says to me this is probably an acid base reaction. But another way you can verify that is you can look at this hydrogen here and we can notice that this hydrogen is going to get transferred because it winds up here. So it's no longer going to be a part of this N. 03 It's now a part of this NH four. So that's a hydrogen transfer, which indicates acid base reaction. All right, The next thing we're gonna do is we're given this other equation that we're going to look at which is NH three plus oh two yields N O plus H 20 And I've spaced this out because I already know that this balancing thing is going to be a lot more convoluted than the one that we had up here. There are no polly atomic ions that we can keep together. So we just have to list each individual element. So we've got nitrogen, hydrogen and oxygen. I'm going to go through these and I'm going to add up what we've got one nitrogen, three hydrogen, two oxygen's on the other side, one nitrogen two hydrogen bombs. And to oxygen's unfortunately, this is not balanced and it's going to take a little while and a lot of just working things out and seeing what works. So the first thing I'm gonna do is I'm going to try to balance these hydrogen by saying we have two of these here and three of these here. Well now I have to update everything. So on this side we have now to nitrogen and six hydrogen On this side, we now have six hydrogen and four oxygen's. As a general rule of thumb when you're doing things like this, don't mess with the oxygen. It'll probably balance itself out. And if not it will have a really easy thing right at the end. But don't mess with that right now. Well this is all well and good except we don't have our nitrogen balance or are oxygen's balance anymore. So I'm going to update this over here and give this one two nitrogen. Well that throws off our oxygen's too. So now we have five of these which is really inconvenient because if it was an even number, we could just put whatever we need to multiply by two here and it would be balanced. But it's not, it's an odd number. So this is what I mean when I say trying things and seeing if they work, um I'm gonna go ahead and tell you that after a lot of guests and check, I worked that we have six of these. So we have to go back and update. We now have 12 hydrogen and eight oxygen's here. Well now are hydrogen czar out of balance. So I'm going to go over here And change this to a four. So now we have four nitrogen, 12 hydrogen two oxygen's well, we have to balance, we have rebalance our nitrogen. So I'm going to have four, Which means we have four oxygen's here and six here, which means we have 10 oxygen's Well now everything is balanced except for the oxygen and we have an even number for the oxygen. So we can just put a five here. That's going to be 10. And now we have our balanced equation. I'm going to write it again over here just so it's cleaner. We have four n. H three Plus. 50.2 yields four and O plus six H 20 Okay, the next thing we're going to do is we're going to find out what the maximum mass of N. O We can produce is if we're given that we have one kg of NH three and to do this, we're going to find the Relative formula mass of those two things that we need. And what that means is first we're gonna find the molar mass of NH three and N. O. And then multiply that by the coefficients that we've got and that's our relative formula mass. So at age three is going to be the atomic mass of Nitrogen which is 14.007. And this you can find on a periodic table plus three times 1.008, which is the atomic mass of hydrogen. When you some that you get 17.031. But remember we have two multiplied by the coefficient we have a four here. So we're going to multiply that by four. And that is going to give us 68 0.12 four. Oh, all right. We're going to do the same thing with N. O. The atomic mass of nitrogen is 14 .007. The atomic mass of oxygen is 15.999. Some of those together you get um 30 .006. And once again, don't forget to multiply by our four. That's going to give us 120 0.2 four. So to find our maximum mass of N. O. We're going to divide or what we found for that By our given value 68.1- four. So that's going to be 1 20 .024, Divided by 68 .1-4 which is equal to um 1.76. And then we're gonna multiply that by R1 kg which is just the same thing. But now it's in kilograms. The next step is we need to figure out how much oxygen we're going to require to run that reaction. So to get our 1.76 kg of n. o, how much oxygen do we need? The process is exactly the same. We're going to find The molar mass. We're going to multiply it by the coefficient and then divide it by the given value 68.1-4. So two times the molecular mass. Sorry, the atomic mass of oxygen is 15 points 999. Which is going to give us 31.998 Multiply that by five is 159 0.99 Um We're going to divide that by 68 .1-4 are given To get 2.3 five kg of 02. And finally we're going to find our percent yield. What % yield is is no reaction that you run is going to be 100 efficient, which means there's always going to be stuff left over that did not react. So we're trying to find the efficiency of this reaction. So so our percent yield is our actual yield, which we're going to be given by our theoretical yield, Which is how much would react if this was 100 efficient. So times 100 Percent to get our percent here are actual is given to us at 1.45 kg of NH three. And our theoretical, sorry, not NH three n. O. And our theoretical is 1.76, So 1.45 Divided by 1.76 Times 100%,, is 82 .39%. And that is our final answer.

Okay. So in this reaction we're starting with a solid. Okay. And we're producing two gases. All right. And we want to figure out what the total pressure is in the system? Well, our gases are in a 1 to 1 ratio. Okay. Search. I so I can say I'm going to form equal amounts of them. I'll call them both X. And since this is a solid, I won't worry about it right now. It's not going to contribute to the partial pressure or be involved in RK expression. So okay equals the partial pressure of NH three times the partial pressure of H. I. And we know okay, is 0.215 So that's just equals X squared. Okay? So X. Is going to equal zero 464 atmospheres. And that's the partial pressure of both of them. So that's your partial pressure of NH three. And it's also the partial pressure of the H. I. So our total pressure, we'll just go ahead and add those together and we'll get 0.9 to a ATM and that will be our total pressure. Okay. And then what we want to do is try to figure out how much of this solid is left when we're done. Okay. So I know I formed 0.464 ATM of ammonia. So I'm gonna use that to find how many moles of ammonia were produced. And when you use the ideal gas law PV equals NRT, I'm gonna rearrange it a little bit for moles. So our pressure is 0.464 and our volume is 75 liters. We'll go ahead and divide that by R which is 0.8 to one. And our temperature in kelvin which is 673. So we know that we produced this many malls of ammonia. Oh well that means that we must have used that same amount 0.630 moles of our solid ammonium iodide. Okay, that was used up. And let's see how many moles of the ammonia. My idea, we start with where we started with 20 g. So we'll do a little sticky amateur here, grams to moles. And this is going to weigh 1 44 0.95 It's a smaller mass. So I started with 0.138 moles. Okay. And we just discovered that we used 0.630 moles. So after the reaction is done, I had 0.75 malls remaining. So I'll just go ahead and multiply that by its molar mass. 1 44 0.95 g per mole. And we'll figure out that we had 10 9 grams of our solid remaining.


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