Chemistry question on excess

[Baller97]

Hey guys

So I was goin over a few past paper questions and got stuck on this one:-

 

Nitrogen monoxide may be removed from industrial emissions via a reaction with ammonia as shown by the equation below.

4NH3(g) + 6NO(g) → 5N2(g) + 6H2O(l)

(iii) 30.0 dm3 of ammonia reacts with 30.0 dm3 of nitrogen monoxide at 100 °C.

Identify which gas is in excess and by how much and calculate the volume of nitrogen produced.

 

Somebody please explain?

Thanks in advance!

[Vioh]

 

Hey guys

So I was goin over a few past paper questions and got stuck on this one:-

 

Nitrogen monoxide may be removed from industrial emissions via a reaction with ammonia as shown by the equation below.

4NH3(g) + 6NO(g) → 5N2(g) + 6H2O(l)

(iii) 30.0 dm3 of ammonia reacts with 30.0 dm3 of nitrogen monoxide at 100 °C.

Identify which gas is in excess and by how much and calculate the volume of nitrogen produced.

 

Somebody please explain?

Thanks in advance!

 

 

First, Avogadro's theory tells you that: Equal volume of gases, at the same temperature and pressure, will contain an equal number of moles. So this means that 30dm3 of NH3 has exactly the same number of moles as 30dm3 of NO does.

 

From the balanced equation, the ratio of NH3 to NO is 4/6 = 2/3 (i.e. if you have 2 moles of NH3, you need at least 3 moles of NO). This means 30dm3 of NH3 will need at least 45dm3 (=30/2*3) of NO to fully react. But the question says that you only have 30dm3 of NO. Therefore, NO must be the limiting reagent, while NH3 is in excess.

 

To calculate the amount of N2 produced, you need to use the number of moles of the limiting reagent. From the balanced equation, the ratio of NO to N2 is 6/5. All you need to do now is to take the number of moles of NO and divide it by this ratio --> 30 / (6/5) = 30*5/6 = 25 dm3

 

So the answer is: NH3 is in excess and 25dm3 of N2 is produced. Feel free too ask if you have further questions. Good luck with the exam!

[Baller97]

 

 

Hey guys

So I was goin over a few past paper questions and got stuck on this one:-

 

Nitrogen monoxide may be removed from industrial emissions via a reaction with ammonia as shown by the equation below.

4NH3(g) + 6NO(g) → 5N2(g) + 6H2O(l)

(iii) 30.0 dm3 of ammonia reacts with 30.0 dm3 of nitrogen monoxide at 100 °C.

Identify which gas is in excess and by how much and calculate the volume of nitrogen produced.

 

Somebody please explain?

Thanks in advance!

 

 

First, Avogadro's theory tells you that: Equal volume of gases, at the same temperature and pressure, will contain an equal number of moles. So this means that 30dm3 of NH3 has exactly the same number of moles as 30dm3 of NO does.

 

From the balanced equation, the ratio of NH3 to NO is 4/6 = 2/3 (i.e. if you have 2 moles of NH3, you need at least 3 moles of NO). This means 30dm3 of NH3 will need at least 45dm3 (=30/2*3) of NO to fully react. But the question says that you only have 30dm3 of NO. Therefore, NO must be the limiting reagent, while NH3 is in excess.

 

To calculate the amount of N2 produced, you need to use the number of moles of the limiting reagent. From the balanced equation, the ratio of NO to N2 is 6/5. All you need to do now is to take the number of moles of NO and divide it by this ratio --> 30 / (6/5) = 30*5/6 = 25 dm3

 

So the answer is: NH3 is in excess and 25dm3 of N2 is produced. Feel free too ask if you have further questions. Good luck with the exam!

 

Hmm... I still don't get how you find the excess and limiting though :/

[Vioh]

 

 

 

Hey guys

So I was goin over a few past paper questions and got stuck on this one:-

 

Nitrogen monoxide may be removed from industrial emissions via a reaction with ammonia as shown by the equation below.

4NH3(g) + 6NO(g) → 5N2(g) + 6H2O(l)

(iii) 30.0 dm3 of ammonia reacts with 30.0 dm3 of nitrogen monoxide at 100 °C.

Identify which gas is in excess and by how much and calculate the volume of nitrogen produced.

 

Somebody please explain?

Thanks in advance!

 

 

First, Avogadro's theory tells you that: Equal volume of gases, at the same temperature and pressure, will contain an equal number of moles. So this means that 30dm3 of NH3 has exactly the same number of moles as 30dm3 of NO does.

 

From the balanced equation, the ratio of NH3 to NO is 4/6 = 2/3 (i.e. if you have 2 moles of NH3, you need at least 3 moles of NO). This means 30dm3 of NH3 will need at least 45dm3 (=30/2*3) of NO to fully react. But the question says that you only have 30dm3 of NO. Therefore, NO must be the limiting reagent, while NH3 is in excess.

 

To calculate the amount of N2 produced, you need to use the number of moles of the limiting reagent. From the balanced equation, the ratio of NO to N2 is 6/5. All you need to do now is to take the number of moles of NO and divide it by this ratio --> 30 / (6/5) = 30*5/6 = 25 dm3

 

So the answer is: NH3 is in excess and 25dm3 of N2 is produced. Feel free too ask if you have further questions. Good luck with the exam!

 

Hmm... I still don't get how you find the excess and limiting though :/

 

 

Right. The balanced equation tells you that:

• If you have 2 moles of NH3, you will need 3 moles of NO for a full reaction (Because 2/3 = 4/6, i.e. the ratios are equal)

Avogadro's theory (which I've stated above) tells that you can directly convert the number of moles into volumes if the reaction happens under the same temperature and pressure. This means that:

• If you have 2dm3 of NH3, you will need 3dm3 of NO for a full reaction (Notice that the ratio is still 2/3)
• --> If you have 30dm3 of NH3, you will need 45dm3 of NO for a full reaction (Because 30/45 = 2/3, i.e. the ratios are still equal)

Now the question tells you that you have 30dm3 of NH3, but only 30dm3 of NO (which is less than 45dm3). This means that NO is the limiting reagent, because you don't have enough of it for a full reaction. And because NO is limiting, then NH3 must be in excess.

Tell me whether this explanation is clear enough

[Baller97]

 

 

 

 

Hey guys

So I was goin over a few past paper questions and got stuck on this one:-

 

Nitrogen monoxide may be removed from industrial emissions via a reaction with ammonia as shown by the equation below.

4NH3(g) + 6NO(g) → 5N2(g) + 6H2O(l)

(iii) 30.0 dm3 of ammonia reacts with 30.0 dm3 of nitrogen monoxide at 100 °C.

Identify which gas is in excess and by how much and calculate the volume of nitrogen produced.

 

Somebody please explain?

Thanks in advance!

 

 

First, Avogadro's theory tells you that: Equal volume of gases, at the same temperature and pressure, will contain an equal number of moles. So this means that 30dm3 of NH3 has exactly the same number of moles as 30dm3 of NO does.

 

From the balanced equation, the ratio of NH3 to NO is 4/6 = 2/3 (i.e. if you have 2 moles of NH3, you need at least 3 moles of NO). This means 30dm3 of NH3 will need at least 45dm3 (=30/2*3) of NO to fully react. But the question says that you only have 30dm3 of NO. Therefore, NO must be the limiting reagent, while NH3 is in excess.

 

To calculate the amount of N2 produced, you need to use the number of moles of the limiting reagent. From the balanced equation, the ratio of NO to N2 is 6/5. All you need to do now is to take the number of moles of NO and divide it by this ratio --> 30 / (6/5) = 30*5/6 = 25 dm3

 

So the answer is: NH3 is in excess and 25dm3 of N2 is produced. Feel free too ask if you have further questions. Good luck with the exam!

 

Hmm... I still don't get how you find the excess and limiting though :/

 

 

Right. The balanced equation tells you that:

• If you have 2 moles of NH3, you will need 3 moles of NO for a full reaction (Because 2/3 = 4/6, i.e. the ratios are equal)

Avogadro's theory (which I've stated above) tells that you can directly convert the number of moles into volumes if the reaction happens under the same temperature and pressure. This means that:

• If you have 2dm3 of NH3, you will need 3dm3 of NO for a full reaction (Notice that the ratio is still 2/3)
• --> If you have 30dm3 of NH3, you will need 45dm3 of NO for a full reaction (Because 30/45 = 2/3, i.e. the ratios are still equal)

Now the question tells you that you have 30dm3 of NH3, but only 30dm3 of NO (which is less than 45dm3). This means that NO is the limiting reagent, because you don't have enough of it for a full reaction. And because NO is limiting, then NH3 must be in excess.

Tell me whether this explanation is clear enough

 

Ahhh yeah got it now. Thank you so much!

[ultimateone]

 

 

Hey guys

So I was goin over a few past paper questions and got stuck on this one:-

 

Nitrogen monoxide may be removed from industrial emissions via a reaction with ammonia as shown by the equation below.

4NH3(g) + 6NO(g) → 5N2(g) + 6H2O(l)

(iii) 30.0 dm3 of ammonia reacts with 30.0 dm3 of nitrogen monoxide at 100 °C.

Identify which gas is in excess and by how much and calculate the volume of nitrogen produced.

 

Somebody please explain?

Thanks in advance!

 

 

First, Avogadro's theory tells you that: Equal volume of gases, at the same temperature and pressure, will contain an equal number of moles. So this means that 30dm3 of NH3 has exactly the same number of moles as 30dm3 of NO does.

 

From the balanced equation, the ratio of NH3 to NO is 4/6 = 2/3 (i.e. if you have 2 moles of NH3, you need at least 3 moles of NO). This means 30dm3 of NH3 will need at least 45dm3 (=30/2*3) of NO to fully react. But the question says that you only have 30dm3 of NO. Therefore, NO must be the limiting reagent, while NH3 is in excess.

 

To calculate the amount of N2 produced, you need to use the number of moles of the limiting reagent. From the balanced equation, the ratio of NO to N2 is 6/5. All you need to do now is to take the number of moles of NO and divide it by this ratio --> 30 / (6/5) = 30*5/6 = 25 dm3

 

So the answer is: NH3 is in excess and 25dm3 of N2 is produced. Feel free too ask if you have further questions. Good luck with the exam!

 

 

Thanks Vioh for your answer, but the question also asks you to mention by how much specifically the gas in excess is in excess. I think you forgot that part. Mind elaborating how to calculate that too?

[IbTrojan]

I think he did that in a post above...