When a sample of NO(g) (388.1 grams) is placed in 290.0 L reaction vessel at 500.0 °C and allowed to come to equilibrium the mixture contains 53.28 grams of N2(g). What is the concentration (mol/L) of O2(g)?
2NO(g) = N2(g)+O2(g)
I would do this.
moles NO = 388.1/molar mass NO = ??
M NO = moles/290L = ??
The problem states that you obtain 53.28 g N2. moles N2 = 53.28/molar mass N2.
Then M N2 = moles/290 L = ??
Since the equation tells you that you have 1 mole O2 and 1 mole N2, you know the two are equal in concns (and moles for that matter).