Wednesday
November 26, 2014

Homework Help: chem.

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

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

Answer this Question

First Name:
School Subject:
Answer:

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 ...

Search
Members