The equilibrium constant for the reaction A2(g)+B2(g)⇌2AB(g) at 100°C is 50. If a one liter flask containing one mole of A2 is connected to a 2 litre flask containing two moles of B2and allowed to react, the amount of AB formed at equilibrium at 100°C is

A2(g)+B2(g)⇌2AB(g) Initial moles                       1          2             0                                          -x         -x        +2x At equilibrium              1-x    2-x         2x Equilibrium constants Kc=[AB]2A2B2                                      KC=2x321-x32-x3                                      50=(2x)2(1-x)(2-x) 50x2-150x+100=0                              x=0.934   Number of moles AB = 2x                                         = 1.868 moles