I need with this equilibrium question. I keep getting a negative concentration for C.

Question

A mixture of gases A, B, and C were placed into a 1.00 L container. 10.00 moles of each of the three gases were used to create the mixture. The mixture was then allowed to reach equilibrium at a temperature of 25.0^°. At equilibrium, the number of moles of B in the container is 12.66 moles. The reaction for the process is:

A(g) + 2B(g) ↔ 3C(g)

Calculate the K_c for the reaction at constant temperature.

Answer 1

A(g) + 2B(g) <==> 3C(g)

Use an ICE table to find the equilibrium concentrations. NOTE however that the moles of B at equilibrium is greater than the starting amount (12.66 > 10.00) telling us that the reaction proceeds to the left (towards reactants)

A(g) + 2B(g) <==> 3C(g)

10.......10..............10.........Initial

x.......+2x..............-3x........Change

10+x..10+2x........10-3x.....Equilibrium

10+2x = 12.66, therefore x = 1.33

Equilibrium concentrations are as follows:

[A] = 10 + 1.33 = 11.33

[B] = 12.66

[C] = 10 - 3.99 = 6.010

K= [C]³ / [A][B]²

K = (6.010)³ /(11.33)(12.66)²

K = 0.1195

(be sure to check my math)