April 20, 2024, 10:00:29 AM
Forum Rules: Read This Before Posting


Topic: Need a little Help  (Read 11407 times)

0 Members and 1 Guest are viewing this topic.

Online Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27652
  • Mole Snacks: +1800/-410
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Need a little Help
« Reply #15 on: February 24, 2008, 04:44:36 PM »
Im not sure I know how to do that.  Do you mean to fully react NH3?

As I told you - forget for now about limiting reagents, excess reagents - it doesn't matter. You need simple stoichiometry - from the reaction equation 1 mol of ammonia reacts with 7/4 mole of oxygen. How much oxygen do you need for a given amount of ammonia?

Quote
K so I figured I have 3135 mols of NH3 and 9143 mols of O2.

:( Something wrong here... Show how you get these numbers.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline Shade3

  • Regular Member
  • ***
  • Posts: 10
  • Mole Snacks: +0/-0
Re: Need a little Help
« Reply #16 on: February 24, 2008, 06:39:51 PM »
Oh ok so I need 4/4 oxygen is what your saying? So 3/4 oxygen mols is in excess?  And for those numbers I used PV = nRT

so for NH3 n = (800atm X 120L) / (0.0821 X 373) = 3135 mol
and for O2 n = (800atm X 200L) / (0.0821 X 373) = 5225 mol then X 7/4 = 9143 mol

I thought the numbers were really high but thats how I got it.  The 0.0821 is my given R value.

Online Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27652
  • Mole Snacks: +1800/-410
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Need a little Help
« Reply #17 on: February 24, 2008, 07:01:27 PM »
Oh ok so I need 4/4 oxygen is what your saying? So 3/4 oxygen mols is in excess?

No. For 1 mole of ammonia you need 7/4 mole of oxygen - thats in the reaction equation (learn here how to read reaction equation). 7/4 mole of oxygen for one mole, 2*7/4 mole of oxygen per two moles of ammonia, 0.5*7/4 mole of oxygen per half mole of ammonia and so on.

Quote
And for those numbers I used PV = nRT

so for NH3 n = (800atm X 120L) / (0.0821 X 373) = 3135 mol
and for O2 n = (800atm X 200L) / (0.0821 X 373) = 5225 mol then X 7/4 = 9143 mol

I thought the numbers were really high but thats how I got it.  The 0.0821 is my given R value.

1. Why 373?

2. When converting volume to number of moles you don't use stoichiometric coefficients! Volume/moles conversion is separate step, stoichiometric calculations is separate step. Both calculations are often necessary to solve the same question, but they can't be mixed.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline Shade3

  • Regular Member
  • ***
  • Posts: 10
  • Mole Snacks: +0/-0
Re: Need a little Help
« Reply #18 on: February 24, 2008, 07:28:29 PM »
I used 373 because its the temperatute (373Kelvin).

And I see what you mean but if the equation is balanced then wouldn't you need all of the 200L of oxygen to react with the 120L of NH3, you wouldn't have too much of either.

I read that page but I don't think I understand what you're trying to tell me. 

Sponsored Links