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N2 + H2 <=> 2NH3 Given in part A: Kp = 1.64x10-4 at 673K and they...

N2 + H2 <=> 2NH3 Given in part A: Kp = 1.64x10-4 at 673K and they asked what is the Kc value, I got 5.38x10-8.

Now they are asking in Part B, with 105 moles N2 and 318 moles H2 in a 400L gas chamber, how much ammonia would be produced?.... I set up an ICE table with the Es equaling: E N2= 105-1x, E H2= 318-3x and E NH3 = +2x

following I Set up this, Kc=1.64x10-4(2x)2 /(105-1x)(318-3x)3 ....I believe to solve for X or NH3 a quadratic formula is to be used now, how would you solve for this problem/ Set it up?

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