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The amount of nitrogen gas present at equilibrium is 0.0039 moles.
Volume of the system = 3500 mL or 3.5 L
Concentration of ammonia at equilibrium = 0.25 mol/3.5 L = 0.07 M
Concentration of hydrogen gas at equilibrium = 0.080 mol//3.5 L = 0.02 M
Equilibrium constant for this reaction = 5.81 x 10^5
The reaction for the formation of ammonia is;
N2 + 3H2 ⇄ 2NH3
So;
K = [NH3]^2/[N2] [H2]^3
K [N2] [H2]^3 = [NH3]^2
[N2] = [NH3]^2/K [H2]^3
[N2] = [ 0.07]^2/ 5.81 x 10^5 × [0.02]^3
[N2] = 0.0011 M
Amount of N2 = 0.0011 M × 3.5 L = 0.0039 moles
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