for the reaction 3A(g)+B(g)=2C(g) at a given temperature,Kc=9.0 . what must be the volume of the flask, if a mixture of 2.0 mole each of A,B and C exist in equilibrium?

Given : 
K_c​ = 9.0 
2 moles each of A, B, C

​To find : 
volume, V = ?

Solution : 
For the reaction , 
$3A\left(g\right) + B\left(g\right) \iff 2C\left(g\right)$
The equilibrium constant ,K_c = $\frac{[C{]}^2}{\left[A{]}^3\right[B]}$ 

concentration = $\frac{\text{number of moles}}{\text{volume}}$ 
where , concentration [C] = [A] =[B] = $\frac{2}{V}$ mol L^-1
$9 = \frac{{\left({\displaystyle \frac{2}{V}}\right)}^2}{{\left(\frac{2}{V}\right)}^3\left(\frac{2}{V}\right)}$
$9 =\frac{V^2}{4}$
V = 6 L
​Volume of the flask is 6 L