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Question 2In a laboratory determination of the empirical formula of tin oxide, Sn,O,, a sample of tin is weighed in a crucible Nitric acid is added and then the mix...

Question

Question 2In a laboratory determination of the empirical formula of tin oxide, Sn,O,, a sample of tin is weighed in a crucible Nitric acid is added and then the mixture is heated, and the reaction proceeds to give Sn O,, NOz and HO The unbalanced reaction equation is shown below. The products are further heated in the crucible t0 obtain the dried tin oxide.Sn (s) + HNO; (aq) Sn 0, (s) + H,O (g) + NO_ (g) The following data is collected:Mass crucible (g- 11.25 Mass Sn and crucible (g) 12.25Mass o

Question 2 In a laboratory determination of the empirical formula of tin oxide, Sn,O,, a sample of tin is weighed in a crucible Nitric acid is added and then the mixture is heated, and the reaction proceeds to give Sn O,, NOz and HO The unbalanced reaction equation is shown below. The products are further heated in the crucible t0 obtain the dried tin oxide. Sn (s) + HNO; (aq) Sn 0, (s) + H,O (g) + NO_ (g) The following data is collected: Mass crucible (g- 11.25 Mass Sn and crucible (g) 12.25 Mass of dried Sn O,and crucible (g 12.52 (a) Determine the mass of Sn and mass of dried Sn 0 , (b) Determine the empirical formula of Sn 0 (c) Explain the effect on the calculated empirical formula that would result from each of the following experimental eTors: The reaction mixture bubbles over the crucible upon reaction of the HNO; (aq) with the Sn (s) . (ii) The tin oxide product is not heated sufficiently to dry it completely: (d) Is the reaction below an oxidation-reduction reaction? Justify your answer. Sn (s) + HNO; (aq) Sn O, HO (g) + NO; (g) (e) Sn,0,(s) has a high melting - point when compared to H,O (s) Explain such an observation using the strength ot the particle torces present In the compounds.



Answers

Tin metal (Sn) and purple iodine (I_2) combine to form orange, solid tin iodide with an unknown formula. $$\text { Sn metal }+\text { solid } \mathrm{I}_{2} \rightarrow \text { solid } \mathrm{Sn}_{x} \mathrm{I}_{y}$$ Weighed quantities of Sn and $\mathrm{I}_{2}$ are combined, where the quantity of Sn is more than is needed to react with all of the iodine. After $\operatorname{Sn}_{x}$ I has been formed, it is isolated by filtration. The mass of excess tin is also determined. The following data were collected: Mass of tin (Sn) in the original mixture $1.056 \mathrm{g}$ Mass of iodine $\left(\mathrm{I}_{2}\right)$ in the original mixture $1.947 \mathrm{g}$ Mass of tin (Sn) recovered after reaction $0.601 \mathrm{g}$ What is the empirical formula of the tin iodide obtained?

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.

Okay, so we have compound one, which has 40 g of metal M we've got 12 g of carbon And 47.96 g of oxygen. So if you divide each of these by their respective Mueller masses, We're gonna get .999 moles of carbon, divide our oxygen by 16. That's gonna give us 2, 9, 9 moles of oxygen. And if we're dealing with a carbonate containing compound, um and our metal is m then let's look at the group to metals and their molar masses. Because realistically it's probably likely that we are looking at calcium, Calcium Solar Masses 40.8. Because if you divide this by it by its smaller mass, you're gonna get .999. And that's the smallest mole. Which means you're dividing everything by the smallest small, so the smallest small matches for carbon and calcium. And so you'll find out that you have 11 and three. And this gives you the sub scripts for the carbonate ion that's in the compound. So it makes sense that calcium is going to be our metal. Mm. Okay, so that means that calcium carbonate is the compound. And calcium carbonate is an ionic compound. Cat ion followed by an anti on and when calcium carbonate decomposes From being heated, it's going to form calcium oxide and carbon dioxide gas, which is a colourless gas. Now, if we're looking at compound number two compound number two has got 69 point 59 g of metal M six point oh nine g of carbon and 24.32 g of oxygen. So if we divide carbon by its smaller mass, which is 12 point oh one, you're gonna get 10.507 moles of carbon, Divide oxygen by its smaller mass, which is 16. And that's gonna give you 1.52 moles of oxygen. And um I would look at the group two medals again, look at the molar masses of all of the different metals. And if we assume that, well, our smallest mole is probably this .507 that's going to give us our carbonate are one carbon and our three oxygen's in the compound, then um if you look at barium, smaller mass, which is 1 37.33, When you divide this out, you're gonna end up with .507 moles. So that matches the smallest small, which we were guessing which was the moles of carbon. So that means you have one barium. So that means that here, you're going to end up with B a C 03, that's an ionic compound as well. Metal M is barium. And when barium carbonate decomposes upon heating, it's going to form barium oxide plus carbon dioxide

But formula of mag nation nitrate iss This balanced reaction Yes, and the three and two plus three vegetable because I wanted to NGO and understood driving force for this reaction is production off among young guests. So you're on Calvert. In the initial massive funded, we just point, don't in 3 g. So let's see the myself and you convert a 20 year old on Why did the muscles and she converted 23 203 and two does extend regulated as point point 293 g minus off. Why so now to go through the massive friendly in and victory and to muscles, NGO is eggs eso 4th 3.3. So Fangio by 24.3 from so far energy on, um, also 103 and two, which is supposed to whine, uh, white off 100.9 g of empty three to divide advice 73 72.9 70. But as we know myself and your plus months Gangotri into is going to a +747 video grams. So, uh, my second shoe is this on myself in G three center is this which is the first of going seven video going for 70 g. So after calculating after her calculating, I have got Hawaii, is this point the 579 so massive Andrian and veteran toe this point verify pictograms. Uh, now because play the most percentage job, I'm J three and two most of family and recreation. 30.58 g Marshall 102 years Things Certify waiting 200.9 g of 103 and two divided US geo 0.9 g of country, which is find there is 0 to 7, only to find out more percentage off the train, too, because I maintain 0.8%. The teapot balanced reaction is three m Jesus, the industry which funds in between two on and now that's a sex literature. When the three over here to determine the limiting reaction will amass, a furnished is 17.3 g. Remember, he is since indifferent to this point program for most well trained took 24 points of your family into the orcs with 27 tingling military and Vanish Street. The 670 industry has to live 2.4 g of vanish since we're providing provided to 2.57 g of finished three Does industry is I am inconvenienced int calculate the mass of fish toe found, uh, there's, uh Those 2.57 g of ministry produces this much from massive victories 0.45 g of h two. Now to calculate the standard and still be changed for the reaction. So which the formula is a trash, not reaction is supposed to. Delta is not off products minus our national powerful. The reactions after putting all the values we got, a Delta ash not off The action is minus three honest 38.7 village.

So this question wants us to determine the empirical formula of tin oxide, which would have formed outside of this crucible and this question, trying kind of trying to confuse you by including the weight of the crucible where your calculations. But if you remove that wait, you can get just the weight of the 10 by itself and the weight of the 10 oxide by itself. So since we know that we have 1.591 grams of tin oxide and we know that it was made using 1.253 grams of 10 we can determine that the remaining 0.3 grams 0.338 grams comes from oxygen, so we can now determine how many moles of 10 and how many moles of oxygen we have. Our number of mole of 10. It's found by dividing our weight of 10 Ingram's by the molecular weight of two, or rather, the atomic weight of 10. Ingram's Permal. We should have 0.106 Mol of Tim, And if you do the same for oxygen, we should find you have 0.211 more of oxygen And if we divide the number of oxygen by a number of mole of Tim, we should find we get approximately to move oxygen for everyone moved in. And this gives us our empirical formula of s and 02


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