Under certain conditions, nitrogen and oxygen react to form dinitrogen oxide. Suppose the mixture of 0.482 mol N2 and 0.933 mol O2 is placed in a reaction vessel of volume 10.0 L and allowed to form N2O at a temperature for which Kc = 2.0×10^-37?

I assume you are looking for the final equilibrium concentrations of this reaction.

First, you must balance the chemical equation in order to get the coefficients needed for the rate reaction:

##2N_2 + O_2 rightleftharpoons 2N_2O##

Then you need to set up an ICE (Initial, Change, Equilibrium) chart to find the factors that need to be solved in the equation for a numerical result:
##N_2 = “0.482 Mole”##, ##O_2 = “0.933 Mole”## ##K_c = 2 xx 10^(-37)##
The volume is irrelevant as long as the molar quantities are known and it is a gas mixture. It is only necessary if you are given solution volumes and need to correct from the standard moles/Liter to actual moles in solution for liquid .

##” ” ” “2N_2″ ” ” ” + ” ” ” “O_2″ ” rightleftharpoons ” ” ” “2N_2O##

##”I” ” ” ” ” ” ” “0.482” ” ” ” ” ” ” ” ” “0.933” ” ” ” ” ” ” ” ” ” “0##
##”C” ” ” ” ” ” ” “(-x)” ” ” ” ” ” ” ” ” “( -0.5x)” ” ” ” ” ” ” ” “(+x)##
##”E” ” ” ” ” (0.482-x)” ” “(0.933-0.5x)” ” ” ” ” ” “(+x)##
(this is why you need to balance the equation, not all of the ##O_2## will be used)
The equilibrium equation is

##K_c = ([“Products”]^A)/([“Reactants”]^B)” “##, where

##A## and ##B## are the coefficients in the balanced chemical equation.
The equation for this system at equilibrium is thus:
##2.0 xx 10^-(37) = ([N_2O]^2)/([N_2]^2 * [O_2])##
Substituting the problem values from the ICE chart:
##2.0 xx 10^(-37) = (x^2)/((0.482-x)^2 * (0.933-0.5x))##
Solve for ##x## to calculate the final concentrations of ##N_2##, ##O_2##, and ##N_2O##.