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Calculate the theoretical yield of Benzil (grams), starting with1.00 grams of Benzoin with excess nitric acid. Show your work....

Question

Calculate the theoretical yield of Benzil (grams), starting with1.00 grams of Benzoin with excess nitric acid. Show your work.

Calculate the theoretical yield of Benzil (grams), starting with 1.00 grams of Benzoin with excess nitric acid. Show your work.



Answers

Consider the reaction for the synthesis of nitric acid: $$3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g)$$ (a) If $12.8 \mathrm{L}$ of $\mathrm{NO}_{2}(g),$ measured at $\mathrm{STP}$, is allowed to react with $14.9 \mathrm{g}$ of water, find the limiting reagent and the theoretical yield of $\mathrm{HNO}_{3}$ in grams. (b) If $14.8 \mathrm{g}$ of $\mathrm{HNO}_{3}$ forms, what is the percent yield?

Hello students in this question we have to calculate the concentration of nitric acid in moles per liter. Okay, moles per liter. In a simple which has a density of 1.41 g per millimeter. And mass percentage of nitric acid is equals to 69%. Okay, so we can calculate that we have given the density. So for 300 ml we can calculate the volume for the 100 grand. We can calculate the volume be so it will be equals 200 g divided by 1.41 That is equals two 70.92 million L. Okay, so number of modes can be calculated from the mass liver by molecular mass. So masses given as 69%, that is 69 g. Can be taken from 100 g. And we have given the molar mass which is equals to the 40 nitric acid. The molar masses 63. So we get a number of moles equals to 1.39 moles. Okay, so these are the number of moles in this given mass. Okay. No, we can calculate the modularity or concentration. So this concentration, this will be equals to the polarity of the solution. So it will be equals to the moles, develop a volume of the solution Manipulated by 1000. Okay, So we can right here that number of moles are 1.09 and volume is 70.9 to curb this 1000. So from here we get the concentration, it is equal to 15.37 moles per liter. So this become the answer for this problem. Okay, thank you.

So we have this problem here. If 90.8 kilograms of 19 year oxide really well, that's a nitric. I said. It's created so nine 0.9 point 88 kilograms of and to, and we got to convert the smallest, so not smoke black. Let's change this into Graham's first man. Multiply this by the morning Mass, which is 14 close 16 times two 46 grams per mole is equal to equal to about 214.783 Morals of to. And since there is 3 to 2 ratio between nitrogen dioxide and nitric acid and there's 214.73 we need to find X, you can set up her proportion here, like so excess equal to this, X is equal to 1 43 0.18 Pete Mose of Nitric acid, which is each of three. And when you convert this into grams, someone supplies by the molar Mass, which is one cost 40 plus three times 16 63 grounds says 9200 29,020 points for for when you multiply this but 0.95 because that's the that's the percentage yield. And this is simply the theoretical yield. Therefore, we get 8500 69.80 to Gramps of a Chanos. Very that's your answer.

Okay, so, let's go ahead and start by writing are balanced equation. We've got some NH three or ammonia gas and we're gonna react that with oxygen gas. We're gonna make some eno gas And some steam. So H 20 gas. So we'll go through now here and we'll bounce that. Okay, looks like we're going to need to about. We're gonna need a four here, six here For here and then a five here. So we've got our equation balance there. So this is a limiting reacting problem. I know, because I've been given the amounts of both react ints. Mm. So, I know I've got this many grams of ammonia And I've got 10 g of oxygen. It's not readily apparent to me which one is our limiting reactant? So, since we're looking for grams of No, I'm just gonna do the problem twice and choose the smaller answer. Okay, we'll see how much, you know, I can make from the amount of ammonia I have, you know, Oh, The 1-17 4 are more ratio is 4-4, And, you know, is 30 .01g. So, if that's our limiting reactant, that tells us I have enough NH 3 to make 13.4 g of n. o. So, let's check out the oxygen. Now, let's see how much, you know, I can make their So 02, two moles of 02 or more ratio. And then getting that Rno again. And our mole ratio here is 4-5 from our equation 1/32. Thanks. So, This tells us we can make 7502g of N O. So, I have enough ammonia to make this much. I know, but I only have enough oxygen to make this much, you know? So, this is the amount of snow that we're going to produce. Okay, And our oxygen is are limiting reactant and RN. H three is in excess. Okay, So let's see if we can find out how much over. NH three is left over. So if all of my oxygen reacts, let's calculate how much ammonia would react with it. Okay, so we'll change grams of co 2, 2 moles of 02. Then we'll change moles of 02 to moles of ammonia. And then we'll go ahead and change moles of ammonia two g. Okay, so again, oxygen weighs 30 to die atomic. Our mole ratio here is 4-5 Ammonia 17.04 mm. So that tells us that when all of the oxygen reacts, 4-6 g of ammonia reacts as well. But I started with quite a bit more than that. I started with 76 And I reacted 4- six. So after the reaction is done, I have 334g of NH three left over. Mm. Okay, In this final step, we'll find our percent yield their percent yield is our actual or experimental yield over our theoretical yield times 100. The amount that we actually made it was given to us as 6.26 g of the n. o. Our theoretical yield that we calculated above was 7502 g of eno. So this gives us a percent yield of 82 0.9%.

Okay, so, let's go ahead and start by writing are balanced equation. We've got some NH three or ammonia gas and we're gonna react that with oxygen gas. We're gonna make some eno gas And some steam. So H 20 gas. So we'll go through now here and we'll bounce that. Okay, looks like we're going to need to about. We're gonna need a four here, six here For here and then a five here. So we've got our equation balance there. So this is a limiting reacting problem. I know, because I've been given the amounts of both react ints. Mm. So, I know I've got this many grams of ammonia And I've got 10 g of oxygen. It's not readily apparent to me which one is our limiting reactant? So, since we're looking for grams of No, I'm just gonna do the problem twice and choose the smaller answer. Okay, we'll see how much, you know, I can make from the amount of ammonia I have, you know, Oh, The 1-17 4 are more ratio is 4-4, And, you know, is 30 .01g. So, if that's our limiting reactant, that tells us I have enough NH 3 to make 13.4 g of n. o. So, let's check out the oxygen. Now, let's see how much, you know, I can make their So 02, two moles of 02 or more ratio. And then getting that Rno again. And our mole ratio here is 4-5 from our equation 1/32. Thanks. So, This tells us we can make 7502g of N O. So, I have enough ammonia to make this much. I know, but I only have enough oxygen to make this much, you know? So, this is the amount of snow that we're going to produce. Okay, And our oxygen is are limiting reactant and RN. H three is in excess. Okay, So let's see if we can find out how much over. NH three is left over. So if all of my oxygen reacts, let's calculate how much ammonia would react with it. Okay, so we'll change grams of co 2, 2 moles of 02. Then we'll change moles of 02 to moles of ammonia. And then we'll go ahead and change moles of ammonia two g. Okay, so again, oxygen weighs 30 to die atomic. Our mole ratio here is 4-5 Ammonia 17.04 mm. So that tells us that when all of the oxygen reacts, 4-6 g of ammonia reacts as well. But I started with quite a bit more than that. I started with 76 And I reacted 4- six. So after the reaction is done, I have 334g of NH three left over. Mm. Okay, In this final step, we'll find our percent yield their percent yield is our actual or experimental yield over our theoretical yield times 100. The amount that we actually made it was given to us as 6.26 g of the n. o. Our theoretical yield that we calculated above was 7502 g of eno. So this gives us a percent yield of 82 0.9%.


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