Posted by Chris on Monday, November 17, 2008 at 11:57pm.
now if i wrote out the equation..
N2 + H2 ==> NH3
then balanced it..
3N2 + H2 ==> 2NH3
could the numbers given possibly just be extra information that is unneeded, and the volume of ammonia simply be 2 * 22.42 because its at STP?
i really would appreciate some help here..
ok i wrote the other coefficients wrong wow its getting too late for this
Write and balance the equation.
N2 + 3H2 ==> 2NH3.
Use PV = nRT to determine n (# mols) for H2 at the conditions listed.
Convert mols H2 to mols NH3 using the coefficients in the balanced equation.
I assume the question is asking for volume of NH3 produced at the non-standard conditions. Use PV = nRT to convert mols to volume at the non-standard conditions. Post your work if you get stuck.
Erm.. ok..
PV = nRT
(2.25)(13.7) = n(.08206)(366.15)
n = 1.026
1.026molH2 * 2molNH3/3molH2 = .684molNH3
PV=nRT
(?)(?)=(.684)(.08206)(?)
Ohh, nevermind, your saying I should use the same values I used before for P and T ok..
what I got was 9.13L NH3
thank you so much for your help!
9.13 L is correct. By the way, there is a much easier way to do this that I forgot to post after doing it the hard way. But th long way ALWAYS works and this short cut works only in circumstances where all of the materials are gaseous at the conditions listed AND we want the final conditions at th same P and T as the initial conditions. When all are gases, we can cut out the mole step and simply use volumes.
13.7 L x (2 mols NH3/3 mols H2) = 13.7 x 2/3 = 9.13 L. Quick, huh?