For $A (g) + B (g) 2 C (g)$ , $K_{c} = 6.25$ at constant temperature. 1.0 mol each of A and B are placed in a 1.0 L vessel and allowed to reach equilibrium. All of C is then removed instantly, and the system re-equilibrates in the same vessel. Calculate the new amount of C, in mol, to two decimal places.

Numerical answer type, answer in mol, tolerance 0.005.

Answer: 0.49 mol

First equilibrium:

4 x^{2} / (1 - x)^{2} = 6.25

2 x / (1 - x) = 2.5

and

x = 0.5556

mol, leaving 0.4444 mol each of A and B. After removing C:

2 y / (0.4444 - y) = 2.5

, giving

y = 0.2469

mol. New C

= 2 y = 0.4938

mol.