[Knight]

Question:
Assume that at 25 C, with AlCl3 as a catalyst, the following equilibrium can be established between the liquids cyclohexane and methylcyclopentane.
C6H12 <=> C5H9CH3HCH3
Kc = 0.143
If initially 1.00 x 10^2 g cyclohexane is present, what mass of methylcyclopentane will be present in the equilibrium mixture? (Hint: Does the volume of solution matter?)

Attempt:
I attempted to use ICE, but since I do not know the concentration (M) for the reactants, I do not know how to proceed.
I could get the moles of cyclohexane, but then don't I need to convert it to M?

Please advice. Thank you in advance.

[kingchemist]

You could use density to find the volume of the cyclohexane. (and density of product) However, I wonder if you have to? Could you assume the total volume will not change very much in the reaction and so will be constant. So moles/volume are equivalent to moles. You know moles of cyclohexane (initial).
This would allow you to set up the ICE table. Let change be -x, so change in product = +x and at equilibrium, cyclohexane = moles - x and product = x
hey

[valdorod]

volumes cancel out for this particular setup

Kc = [C5H9CH3HCH3]/[C6H12]          concentration is moles over volume and the reaction is taking place in the same volume thus....

kc = (mol C5H9CH3HCH3/Vol)/(mol C6H12/Vol)      vol cancels and you end up with

Kc = (mol C5H9CH3HCH3)/(mol C6H12l) 

keep in mind that whis works only because the total exponents on the numerator = total exponents on denominator
hey