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.