[qbkr21]

I missed this question and I need your help.  Since the temperature and pressures were equal I assumed each had an equal number of molecules and I would just divide the ratios.  I tried 2.8/8.4 and got (1/3) or .33333333333
I am looking for the correct answer in units: Liters.

Thanks,
qbkr21

[macman104]

Right, temperature and pressure are the same, so the only thing that is different is volume, which will indicate moles.  So, we can make our lives easy and assume that at this temperature and pressure, 1 mole occupies a volume of 2.8L.  If that is the case, can you solve the problem now (hint, convert L to mole, and then find moles of NH₃, then convert back to L).

[vhpk]

I missed this question and I need your help.  Since the temperature and pressures were equal I assumed each had an equal number of molecules and I would just divide the ratios.  I tried 2.8/8.4 and got (1/3) or .33333333333
I am looking for the correct answer in units: Liters.

Thanks,
qbkr21

No, you can't convert to mole since you don't know about temperature or pressure, the thing you can only use here is the molar ratio is equal to the volume ratio in the definite temperature and pressure.
So first, you must compare which compound reacts completely and then basing on that compound for calculation. And you can infer directly the volume of ammonia but not depending on its moles

[qbkr21]

 

[macman104]

Looks good to me (didn't check the math, but the setup is correct).  To be absolutely correct, you should add another dimension there that is:

1 mol N₂ / 1 L N₂

But it's just 1/1, so it doesn't affect the calculation, but right now your units don't cancel.

[qbkr21]

Thanks for the HELP.  Oh the wonders of the Internet!

[sjb]

Sorry, but I have to disagree here.

If you have 2.8 litres of N₂, and 8.4 litres of H₂ forming NH₃ (completely) then neither reactant is in excess. As the temperature and pressure are the same, the only change is the number or moles, which is directly proportional to the volume.

So 1 N₂  2 NH₃; and
2.8 litres N₂  ²/₁ x 2.8 litres NH₃

which means how much NH₃ ?

S

[macman104]

Whoops, I made one slight math error, setup is correct, but my math was wrong (just wasn't thinking about it).  1 mol N₂ is not 1 L, it's 2.82 L.  So if you do that, then mine and sjb answers should be the same.

[qbkr21]

Sorry, but I have to disagree here.

If you have 2.8 litres of N₂, and 8.4 litres of H₂ forming NH₃ (completely) then neither reactant is in excess. As the temperature and pressure are the same, the only change is the number or moles, which is directly proportional to the volume.

So 1 N₂  2 NH₃; and
2.8 litres N₂  ²/₁ x 2.8 litres NH₃

which means how much NH₃ ?

S

Thanks 5.6

Could I have used the 8.4 L of H_2 gas?  Do the Coefficient for the products go on top?  Suppose I used the 8.4

8.4 L H_2 * (2/3) = 5.6

[sjb]

Thanks 5.6

5.6 what? pints (US or UK), metres, ounces. ... ?

(don't forget your units!)

Could I have used the 8.4 L of H₂ gas?  Do the Coefficient for the products go on top?  Suppose I used the 8.4

8.4 L H₂ * (2/3) = 5.6

Yes, you could, in fact it's a good way to check if you're not sure which is in excess. Say for instance you had only 8.1 litres of H₂. By this approach you'd see that you could only make 5.4 litres of ammonia, and so hydrogen is the limiting reagent.