Question

The first step in the Ostwald process for producing nitric acid is as follows:
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
If the reaction of 0.880 mol of NH3 and 0.469 mol of O2 yields 0.290 mol NO, what is the percent yield of this reaction?

Answer 1

Do they specify which reagent or product you should find the yield?

Answer 2

No. What I posted was all that was given.

Answer 3

Ok. As they say ''yields NO''. Then I will assume it is this product.
Again, we work on find the limiting and excess reagents.
The molar ratio of NH3 to O2 is 4:5 or 1:1.25.
If you are given 0.88mol of NH3, you are expected to have \[\frac{ 0.88*1.25 }{ 1 }=1.1mol\] of O2, but you are given less, which is 0.469mol.
Therefore O2 is the limiting reagent and NH3 is in excess. Hence we work out between the limiting and the NO.
The molar ratio of O2 to NO is 5:4 or 1:0.8
If we have 0.469 mol of O2, we are expected to have\[\frac{ 0.469*0.8 }{ 1 }=0.3752mol\] of NO, but we are given with 0.29mol of it.

Ultimately, to find percentage yield, we use\[\frac{ Obtained yield*100 }{ expected yield }=\frac{ 0.29*100 }{ 0.3752 }=77.29%\].
The percentage yield is then 77.29%

Answer 4

Thanks!

Answer 5

Thanks for explaining that. I have another question that is similar to this one but they gave grams instead of moles.

Answer 6

Are you able to help me with it?

Answer 7

Well, as I explained before, instead of working with ratio of moles, work with ratio of masses.
Try it out then show me your working.

Answer 8

What is the percent yield of chromium when 12.7 g of Cr is produced from the reaction of 40.0 g of Cr2O3 with 8.00 g of Al according to the following balanced chemical equation?
2Al + Cr2O3 → Al2O3 + 2Cr

Answer 9

That's the question by the way.

Answer 10

Cr2O3 has a mass of 152g

Answer 11

The 12.7g Cr is the actual

Answer 12

First, basing from the reactants side, find the limiting and excess reagents.

Answer 13

Do you find that by doing 40.0g Cr2O3 * (2 mol Al/1 mol Cr2O3)?

Answer 14

No. Do not mix moles with masses.
Follow my steps.

Answer 15

First consider the Mr of the reactants.

Answer 16

Al is 27g
Cr2O3 is 152g, like you told me.

Answer 17

According to the reaction, (2*27) or 54g of Al are to 152g of Cr2O3, right?

Answer 18

ya

Answer 19

If you have 8g of Al, it is expected for it to react with \[\frac{ 8*152 }{ 54 }=22.5g \] of Cr2O3, but you are given 40.0g of it. Can you tell me now which is the limiting and which is in excess?

Answer 20

It would be Al since there is only 22.5g of it compared to the 40.0g of Cr2O3

Answer 21

Correct!
So now we work with the limiting reagent ratio to that of Cr.
Try this step out. I will be back!

Answer 22

Well? Finding hard?

Answer 23

Cr2O3 = 22.535
Cr = 15.418

Answer 24

Or did I do that completely wrong?

Answer 25

Because I did (8.00g Cr2O3)(mol Al)(1 mol Cr2O3)(152g Cr2O3)/(26.98g Al)(2 mol Al)( mol Cr2O3) = 22.535

Answer 26

And (8.00g Al)(mol Al)(2 mol Cr)(52g)/(26.98g Al)(2 mol Al)( mol Cr) = 15.418

Answer 27

Nice one!

Answer 28

So that's all right?

Answer 29

Finally use the formula of percentage yield that I gave previously and work out the percentage.

Answer 30

Yes, you got it!
Now towards the final step. Find the percentage.

Answer 31

The 15.418 is the theo right?

Answer 32

Since the problem is looking for Cr

Answer 33

Yes. That will be the expected yield.

Answer 34

So (12.7)/(15.418) *100 = 82.37%

Answer 35

That's it. You got it.

Answer 36

Thank you so much for your help! You're so awesome!

Answer 37

My pleasure.