[jennielynn_1980]

I can't figure out how to do this question.  I can't figure out what I need to know in order to solve it.

Here is the question and where I have gotten so far:

Consider the following reaction, which takes place in an autoclave at 250 degree C adn 800 atm.

NH₃ (g) + 7/4 O₂ (g) -----> NO₂ (g) + 3/2 H₂O (g)

Into the reaction vessel has been placed 200L of NH₃(g) and 120L of O₂ (g).  The reaction is allowed to go to completion.  Determine the quantity in moles of the gas that remains unreacted.


So I figured I need to know how many moles of NH3 and O2 there are so using n=PV/RT  I got:
NH₃ - 3726.3 mol
O₂ - 2235.8 mol

That takes care of the moles that I am starting with right?  Now do I compare those to the number of moles of NO₂ and H₂O by doing a volume:volume comparison?

[Dan]

You're on the right track,

How many moles of ammonia react with 1 mol of oxygen?
So how many moles of ammonia react with 2235.8 mol of oxygen?
How many moles of ammonia are left over?

[Borek]

So I figured I need to know how many moles of NH3 and O2 there are so using n=PV/RT  I got:
NH₃ - 3726.3 mol
O₂ - 2235.8 mol

While you are correct that you should start with calculation of number of moles of both reactants, both these values are probably wrong (although you have calculated them correctly ). My take is that you should assume both volumes were measured at STP (whatever it means) and 800 atm/250 C is only information about the condition required for the reaction.

You know reaction equation, you know amounts of substances, all you have to do now is to calculate limiting reagent.

Oops, Dan beat me Mostly because I was checking where did you get these amount of moles

[anarchron]

I would preferrably balance the equation so as to not have fractions. It makes scaling much easier:

4NH3 + 5O2 -> 4NO + 6H2O

Remember when you calculatate the remainder NH3 that you scale up the remainder appropriately ie:

3726.3/4 - 2235.8/5 = 4x

x = remainder NH3

[jennielynn_1980]

So I figured I need to know how many moles of NH3 and O2 there are so using n=PV/RT  I got:
NH₃ - 3726.3 mol
O₂ - 2235.8 mol

While you are correct that you should start with calculation of number of moles of both reactants, both these values are probably wrong (although you have calculated them correctly ). My take is that you should assume both volumes were measured at STP (whatever it means) and 800 atm/250 C is only information about the condition required for the reaction.

You know reaction equation, you know amounts of substances, all you have to do now is to calculate limiting reagent.

Oops, Dan beat me Mostly because I was checking where did you get these amount of moles

I can see how that info might be only about the conditions required for the experiment.  Just curious though, how would it affect my answer to assume standard temp and pressure?  Would it give the same percentage of gas in excess?  I didn't even think of doing it this way so now that you suggested it I am interested  

I think I already figured out that answer using the values I originally posted.

1 mole of NH3 required 7/4 mole of O2 for reaction.
Therefore, 3726.3 moles of NH3 requires 7/4 x 3726.3 mole of O2 = 6521.0 mole O2

1 mole of O2 = 4/7 mole NH3
1 mole o2 = 4/7 moles of NH3

2235.8 mole of O2 = 4/7 x 2235.8 = 1277.6 mole of NH3

3726.3 - 1277.6 = 2448.7 moles unreacted NH3

Therefore 2448.7 moles of NH3 remain unreacted.

[Borek]

I can see how that info might be only about the conditions required for the experiment.  Just curious though, how would it affect my answer to assume standard temp and pressure?  Would it give the same percentage of gas in excess?  I didn't even think of doing it this way so now that you suggested it I am interested

You are right - same percentage in excess.

Therefore 2448.7 moles of NH3 remain unreacted.

Correct.

Now that you know how to calculate by yourself, you may save our time checking results like I do - with my EBAS I am too lazy to calculate all these things manually