At a certain temperature the equilibrium constant, kc, equals 0.11 for the reaction: 2 icl(g)? i2(g) + cl2(g). what is the equilibrium concentration of icl if 0.75 mol of i2 and 0.75 mol of cl2 are initially mixed in a 2.0-l flask?

For equilibrium reactions, it would be best to use the ICE method (Initial-Change-Equilibrium). Then, let x be the number of moles that reactedto form ICl. 

               2ICl  ------>  I2  + Cl2

I                0              0.75  0.75
C            +2x              -x      -x
--------------------------------------------
E              2x           0.75 -x  0.75 -x

Kc = [I2][Cl2]/[ICl2]^2 = 0.11, the [I2] represents concentration of I2 in mol/L at equilibrium


 
x = 0.451 moles

Thus [ICl] = (2x/2) = x = 0.451 mol/L

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Adozen is 1221/12 = 1.751.75 x 12 = 21therefore, 1.75 is the answer.