chem.

Hey this is my question:

At a temperature of 300°C and a pressure of 40.5 MPa, 90.0 mol of H2(g) and
80.0 mol of N2(g) are injected into a reaction vessel. When equilibrium is
established, 37.0 mol of NH3(g) are present. The number of moles of H2(g) present
in this equilibrium mixture is

This is my equation so far. How do I go from here?
N2(g) + 3H2(g) >> 2 NH3(g)
initially: 80.0 mol 90.0 mol 0
at equilib:80.0 – x 90.0 – 3x 0 + 2x
0+2x=37.0 mol given


So solve for x and use for 90-3x = ??


ok, I got 20 as the H2 (g) present, would this be right?


Hardly.
If 2x = 37, then x = 37/2 =18.5
Then 90-3x=90-3(18.5)=90-55.5 = 34.5. Check my thinking. Check my math.


mmmmmmmmmmm the answer is 45+897x99=11435%19=897

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asked by yo!

  1. Set up your RICE or ICE table;
    (R) 3H2(g) + N2(g) <-> 2NH3(g)
    (I) 90 80 0
    (C) 90-3x 80-x +2x
    (E) 90-3x 80-x 20
    Looking at NH3(g) you can calculate that x = 10, therefore 90-3(10)=60. 60M or (60mol/L) is the answer. I hope that formatting sticks and this makes sense.

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    posted by Bekk

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