Sunday

November 23, 2014

November 23, 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

**Answer this Question**

**Related Questions**

Chemistry - srry I placed the question in the wrong spot. The equation is 2NH3 &...

AP Chem - At a particular temperature, K = 1.00 102 for the following reaction. ...

AP Chem - The energy change H associated with the reaction NBr3(g) + 3H2O(g) -&...

AP Chemistry - The equilibrium constant for thermal dissociation of F2 F2(g)<...

Chem. thankss!!! - 1.0 mol of water (c=4.184 J/g.oC), 1.0 mol of ethanol (c=2.46...

chemistry - How many mol of Argon is required to exert a pressure of 1.00 MPa at...

Chemistry - Calculate the partial pressures and total pressure (in atm) exerted ...

Chem - What is the pressure in a 2.00 L container which holds 0.100 mol of ...

physics - Use R = 8.2 × 10–5 m3 atm/mol K and NA = 6.02 × 1023 mol–1. The ...

CHEM - Which of the following mixtures will result in the formation of a buffer ...