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###### Alevel Chemistry - Moles

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1 year ago

A chemical company receives an order to supply 1.96x10^10dm3 of ammonia at room temperature and pressure. The Haber process produces a 95.0% yield. Calculate the mass of hydrogen, in tonnes, required to produce the ammonia. Give your answer to 3 significant figures.

Hi, so I have gotten as far as working out the moles of NH3 to be 816666666.7mol and I think the mol ratio of hydrogen to ammonia is 3:2 so I have the moles of H2 as 1225000000mol, not sure if it's right, but also not sure where to go from here, please help me, thank you!

Hi, so I have gotten as far as working out the moles of NH3 to be 816666666.7mol and I think the mol ratio of hydrogen to ammonia is 3:2 so I have the moles of H2 as 1225000000mol, not sure if it's right, but also not sure where to go from here, please help me, thank you!

Reply 1

1 year ago

Original post by charlotte05x

A chemical company receives an order to supply 1.96x10^10dm3 of ammonia at room temperature and pressure. The Haber process produces a 95.0% yield. Calculate the mass of hydrogen, in tonnes, required to produce the ammonia. Give your answer to 3 significant figures.

Hi, so I have gotten as far as working out the moles of NH3 to be 816666666.7mol and I think the mol ratio of hydrogen to ammonia is 3:2 so I have the moles of H2 as 1225000000mol, not sure if it's right, but also not sure where to go from here, please help me, thank you!

Hi, so I have gotten as far as working out the moles of NH3 to be 816666666.7mol and I think the mol ratio of hydrogen to ammonia is 3:2 so I have the moles of H2 as 1225000000mol, not sure if it's right, but also not sure where to go from here, please help me, thank you!

The reaction is 95% efficient, so that means 0.95x = 816666666.7, where x is the theoretical yield of ammonia. How might you find x?

Now using this value of x, find the moles of hydrogen using the ratio H2:NH3 = 3:2.

Reply 2

1 year ago

Original post by TypicalNerd

The reaction is 95% efficient, so that means 0.95x = 816666666.7, where x is the theoretical yield of ammonia. How might you find x?

Now using this value of x, find the moles of hydrogen using the ratio H2:NH3 = 3:2.

Now using this value of x, find the moles of hydrogen using the ratio H2:NH3 = 3:2.

Oh, so how you've explained it makes sense so x = 859649122.8mol so moles of H2 using 3:2 is 1289473684mol? But how did you know 0.95x=816666666.7 when the equation is percentage yield = actual yield/theoretical yield ? Sorry it's just confusing me a bit

Reply 3

1 year ago

Original post by charlotte05x

Oh, so how you've explained it makes sense so x = 859649122.8mol so moles of H2 using 3:2 is 1289473684mol? But how did you know 0.95x=816666666.7 when the equation is percentage yield = actual yield/theoretical yield ? Sorry it's just confusing me a bit

The equation is actually:

Percentage yield = actual/theoretical * 100

If we define the actual as 816666666.7, the theoretical as x and use the fact that the percentage yield is 95%:

95 = 816666666.7/x * 100

Dividing both sides by 100:

0.95 = 816666666.7/x

Multiplying both sides by x:

0.95x = 816666666.7

Edit: the moles of H2 you calculated look right. Moles = (mass in grams)/(relative formula mass). Can you now rearrange this to find the mass of hydrogen in grams? Using the fact that 1 tonne is 1,000,000 g, you should now be able to get an answer.

(edited 1 year ago)

Reply 4

1 year ago

Original post by TypicalNerd

The equation is actually:

Percentage yield = actual/theoretical * 100

If we define the actual as 816666666.7, the theoretical as x and use the fact that the percentage yield is 95%:

95 = 816666666.7/x * 100

Dividing both sides by 100:

0.95 = 816666666.7/x

Multiplying both sides by x:

0.95x = 816666666.7

Edit: the moles of H2 you calculated look right. Moles = (mass in grams)/(relative formula mass). Can you now rearrange this to find the mass of hydrogen in grams? Using the fact that 1 tonne is 1,000,000 g, you should now be able to get an answer.

Percentage yield = actual/theoretical * 100

If we define the actual as 816666666.7, the theoretical as x and use the fact that the percentage yield is 95%:

95 = 816666666.7/x * 100

Dividing both sides by 100:

0.95 = 816666666.7/x

Multiplying both sides by x:

0.95x = 816666666.7

Edit: the moles of H2 you calculated look right. Moles = (mass in grams)/(relative formula mass). Can you now rearrange this to find the mass of hydrogen in grams? Using the fact that 1 tonne is 1,000,000 g, you should now be able to get an answer.

Thank you so much! So using the mass/Mr I got 257894736 then divided by 1,000,000 for tones giving me 258tonnes to 3sf, thank you so so much for your help I appreciate it, I was stuck on this question for ages.

Reply 5

1 year ago

Original post by charlotte05x

Thank you so much! So using the mass/Mr I got 257894736 then divided by 1,000,000 for tones giving me 258tonnes to 3sf, thank you so so much for your help I appreciate it, I was stuck on this question for ages.

Shouldn’t it be 2580 and the moles be 2578947368?

Either way, the digits look about right, so I’d assume you’d get most of the marks (or all of them if I’m the one who has made an error).

(edited 1 year ago)

Reply 6

1 year ago

Original post by TypicalNerd

Shouldn’t it be 2580 and the moles be 2578947368?

Either way, the digits look about right, so I’d assume you’d get most of the marks (or all of them if I’m the one who has made an error).

Either way, the digits look about right, so I’d assume you’d get most of the marks (or all of them if I’m the one who has made an error).

No, yes, just recalculated you’re right 😁thanks again!

Reply 7

1 year ago

Original post by TypicalNerd

Shouldn’t it be 2580 and the moles be 2578947368?

Either way, the digits look about right, so I’d assume you’d get most of the marks (or all of them if I’m the one who has made an error).

Either way, the digits look about right, so I’d assume you’d get most of the marks (or all of them if I’m the one who has made an error).

Actually, I’m just overthinking it now, why is the 816666666.7tonnes the actual and not theoretical if it’s the 95% yield so not 100%? would it not be the theoretical and we’re supposed to work out actual?

Reply 8

1 year ago

Original post by charlotte05x

Actually, I’m just overthinking it now, why is the 816666666.7tonnes the actual and not theoretical if it’s the 95% yield so not 100%? would it not be the theoretical and we’re supposed to work out actual?

Theoretical yield = the absolute maximum amount you get out of it

Actual yield = What you actually get. This should be be the basis of your calculations, so you want to aim for an actual yield of 1.96 x 10^10 dm^3 of ammonia. This is why the 8166666666.7 mol is the actual and not the theoretical.

Reply 9

1 year ago

Right, so if it produced 100% THEN i would use theoretical to get the actual?

Reply 10

1 year ago

Original post by charlotte05x

Right, so if it produced 100% THEN i would use theoretical to get the actual?

If it was 100% efficient (which, for future reference can be assumed in questions where the efficiency isn’t stated), then the theoretical = the actual yield. So if this is what you mean, then yes.

Reply 11

1 year ago

Makes sense so when would you use theoretical?Only when the yield isn’t mentioned?

Reply 12

1 year ago

Original post by charlotte05x

Makes sense so when would you use theoretical?Only when the yield isn’t mentioned?

Pretty much.

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