Calculate equilibrium constant?

[Aafia]

Homework Statement

[/B]
The following reaction was allowed to reach the state of equilibrium
2A+B <====> C
The initial amounts of reactant present in one litre of solution were 0.5 mole of A and 0.6 mole of B. At equilibrium the amounts were 0.2 mole of A and 0.45 mole of B and 0.15 mole of C. Calculate equilibrium constant

Homework Equations

Kc= [product]/[reactants]

The Attempt at a Solution

[/B]
Kc= [0.15]/0.2^2×0.45
Kc=0.15/0.018
= 8.3
In my book one of the option of mcq given is 8.3 but the option marked as correct is 11.7 which one is right? If 11.7 is correct then how?

 

[epenguin]

Homework Statement

[/B]
The following reaction was allowed to reach the state of equilibrium
2A+B <====> C
The initial amounts of reactant present in one litre of solution were 0.5 mole of A and 0.6 mole of B. At equilibrium the amounts were 0.2 mole of A and 0.45 mole of B and 0.15 mole of C. Calculate equilibrium constant

Homework Equations

Kc= [product]/[reactants]

The Attempt at a Solution

[/B]
Kc= [0.15]/0.2^2×0.45
Kc=0.15/0.018
= 8.3
In my book one of the option of mcq given is 8.3 but the option marked as correct is 11.7 which one is right? If 11.7 is correct then how?

I do not see any reason your answer be wrong.

The question contains the superfluous information of initial concentrations, but as you can check these are consistent with the final.