For the reaction

Question

NO2(g) + NO(g) N2O(g) + O2(g)
Kc = 0.914. Equal amounts of NO and NO2 are to be placed into a 5.00 L container until the N2O concentration at equilibrium is 0.0446 M. How many moles each of NO and NO2 must be placed into the container?

Answer 1

......NO2 + NO ==> N2O + O2
I......x.....x.....0......0
C.....-y.....-y....y......y
E.....x-y...x-y....y......y

So you want (N2O) = (O2) = 0.0446M
mols = 0.0446*5L = 0.223 = y
Kc = (N2O)(O2)/(NO2)(NO)
Kc = (x-0.223)(x-0.223)/(0.223)(0.223)
Solve for x = mols NO2 = mols NO

Then (NO2) = (NO) = mols/L = mols from above/5L = ?