For the nitrogen fixation reaction 3H2(g) N2(g)<-->2NH3(g) Kc= 6.0 × 10–2at 500°C.If 0.250 M H2and 0.050 M NH3are present at equilibrium what is the equilibrium concentration of N2?I understand how to set up the ICE tableIm just stuck on the algebra solving for X6.0x10^-2 = (0.050 2x)^2 / (0.250-3x)^3 (x)Could you please explain but organized .
