Thursday

October 30, 2014

October 30, 2014

Posted by **yo!** on Sunday, June 18, 2006 at 6:11pm.

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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