In this problem, we are asked to consider burning magnesium in air. One product on burning magnesium air is the product of magnesium and oxygen, and that product is magnesium oxide. The second product is the reaction of magnesium and nitrogen. And that problem problem or that product is magnesium nitrite. Excuse me. When waters met um, added to magnesium nitride, it reacts to form magnesium oxide and ammonia gas. Okay, so here's all of our equations. I'm gonna go ahead and balances. I think we have to do these later, but I can't take not having imbalanced and here we have three, two and a three. I think that should be good. Okay, so part A says based on the charge of the nitride ion, right? The formula for magnesium nitride, which I already did magnesium is a two plus nitride is a three minus. So the formula is M G three N two, part B. So it's right. The balanced equation for already did this here is the balanced equation for part B. What is the driving force for this reaction? Looking at this, I'm gonna say the driving force for this reaction is probably the formation uh and H three, which is a gas. I'm gonna say that that's going to be a driving force for this reaction. And then see, I believe is a long one. So I'm gonna get started with C. On another page. See might actually take two pages. So in an experiment, a piece of magnesium ribbon is burned in air. So we're burning magnesium ribbon. The mass of magnesium oxide and magnesium nitride is 0.470 g after the reaction. And this is a mixture of M G. O. And MG three and two water is added to the crucible. We're all finished, we add water, we heat it to dryness and we get 0.4 86 grams of M G. O. Only. What's the mass percentage of M G three and two in the 0.470 g. That's what we're going to be looking for. Okay, so let's figure this out. I want to get to all of the different things here. So I set this up by a bunch of numbered steps. I'm just gonna go ahead and number of my steps one. My initial mass of magnesium is going to be equal to magnesium oxide mass times. I'm going to call this the molar mass of magnesium divided by the molar mass of M. G. O. Substitute our values in here and we get magnesium. I'm going to call it initial mass. Just previous initial equals 0.486 g times the molar mass of magnesium is 24 31 g in the molar mass of magnesium oxide is 40.31 g per mole. When I do my work here, I got 0.293 g. And that was my mass of magnesium and we'll need that further on. Any problem. For the second thing I got ready, I'm gonna say, let X equal the mass of the mg mass in M G. O. And let y equal the mg mass in MG three and two. We know that X plus Y has to eat my total mass of my product, which we said was 0.470 That's grams. My bed zero 0.293 g of MG. I'm gonna tell you a little later, we're gonna solve this for X. So I'm just gonna go ahead and write X equals 02 93 g minus why? Let's do our third step, which my third step was. Find the masses of M. G. O. And M. G. Three and two. So my M. G. O. Mass magnesium oxide mounts. I just took that equal to acts from above times the molar mass of magnesium oxide divided by the molar mass of magnesium. So I'm gonna go ahead and just write those down. That was 40 0.31 g and 24.31 g. Actually grams promote. But those cancel out. And my mm. G three and two maths. Well why same thing here? Except it's smaller mass of M. G. Three N two and the molar mass of mg times three. So in this case that is 100 point nine. Mhm. I think I wrote 9 to 9 divided by three times 24.31. Okay, if I add for number four I'll keep the same color. If I add 38 plus three B. And I'm gonna go back moment, let's call this three A. And three B. If I add those two, I'm going to get my mass of product which we were given. So I'm just going to rewrite this quickly here. Um massive product equals my NGO mass which is X times 40 point 31/24 40.0.31 plus. Why times 100 0.9 to nine times three times 24.31. Now back over here I have this that we were going to substitute in for X. So let's go ahead and substitute this in for X. I'll switch colors and I'm also going to figure out these values. My mask. My product was 0470 g. We're in a substitute for X 0.293 g. Whoops minus X minus why? Sorry about that. Times this number extended is 1.658 plus Y. Times this extended is 1384 So doing my math here. And I got after I I guess I'll just write about. I have a room here 0.4858 So I did negative 0.0 158 equals negative zero point 274 Why? And I got a Y. Of 0577 g. And this is grams of magnesium in MG three and two. So this is how many grams I have at M. G three and two. Just one more quick step before we can calculate our percentage. Let's figure out our mass of M G three N two. It's going to be equal to 0.57 seven g. Time is the molar mass of this will be again three times 24 31 g per mole and 100 0.99. Do the math here. And I got 0079 nine grams of M G three and two. And last but not least. We're going to take our 0.799 g magnesium nitride divided by the total mass of our sample. Of course, times 100 because it's percent And I got 17.0%. Mg three M two. Okay, next. Yes, there's announced that was part C. Part D. Asked us to um write a balanced chemical equation and do a quick strike geometry. Super easy. So the equation we're given is we're not given the whole equation but I've balanced it for you. We're told that magnesium reacts with ammonia gas at high temperatures. The product is products are magnesium nitrate. Perhaps that's an S. And hydrogen gas. Okay, so I'll need a three right there. Let me shoot 3213 Okay. And were given that we have 63 g of magnesium. We have 2.57 g of ammonia. I'm going to divide each of these by their molar masses to give moles because I like to have balls and I get 0.259 most of magnesium and 0.151 moles of hydrogen. Or excuse me, ammonia. We're asked to find which is limiting. And then we're also asked to find grams of H two produced. Let's do it. I just wanna is never going to do this. But I decided I'm going to I'm going to do something called the mold box here. And for the mailbox, we're going to write down what we're given. I was given magnesium and I was given ammonia in the mold box. We put coefficients first we had three and two. How many moles we have, which I just calculated zero point 259 and 0.151 How many moles we need. And then we'll figure excess wall boxes are sort of fun to find the number that goes right here. We start with this, multiply it by three and divide by two. So we take 0.151 moles of n H three. Figure out what we need. We put our two moles of n H three in the denominator and our three moles of magnesium on the top. So mathematically you can see what we're doing there. And I got zero point two 265 which means I have an excess of 0.0 2 to 5. Then I'm going to switch colors to find the number that goes here. We start multiply it and divide. So we take the same thing here. But we're gonna switch this switch this factor and I got 0.173 I don't have enough. So this is my limiting reactant. Then next we are going to do a simple story geometry. I'm going to start with the moles of what my limiting is do simple strike geometry using my mole ratio for hydrogen and ammonia, which was two and three, which were two and three and my molar mass for hydrogen. And when we're done, I went to three decimal places because I had three there and I got 0.458 g of hydrogen. And then last but not least e any were given that the delta H. It's not a very nice delta. The heat of formation for magnesium nitride is negative for 61 0.8 Kill jobs per mole. Using that equation that we just wrote down. I'll write it down again here and I better put states on since we're doing these. And I'm gonna write down the anthem peas for each one of these. This one we were given as negative 4 61 08 And this is in colegiales promote and I looked up ammonia and got negative 46 19 Remember that Hess's law is products minus reactant. So for the reaction, I'm going to get no coefficient there. So I'm just gonna write one times negative 4 61.8 I'm sorry please, I can't start this over again. I can't start this over again. Whoa. Uh huh. No. And I got negative 3 68.70 killer jewels per mall.