When a sample of NO2(g) (849.9 grams) is placed in 390.0 L reaction vessel at 961.0 K and allowed to come to equilibrium the mixture contains 248.2 grams of NO(g). What is the concentration (mol/L) of NO2(g)?

Question

2NO2(g) = 2NO(g)+O2(g)

I keep getting .0212 or .04736. This is what I'm doing [849.9/molar mass of no2]/83.00L but its not the right answer. The right answer is 0.02616.

Thanks

Answer 1

I assume you want the NO2 at equilibrium although you don't say that in the post.

initial:
NO2 = 849.9/(46*390) = ?? moles/L
NO = 0
O2 = 0

equilibrium:
O2 = not needed.
NO = 248.2/(30*390) = ?? moles/L
NO2 = -----
Since 2 mols NO are produce for each 2 moles NO2 initially, those molecules are 1:1 so subtract NO2 initial - NO equilibrium = NO2 equilibrium.
I did the math and have
NO2 initial = 0.04737 M
NO equilibrium = 0.02121 M
0.04737-0.02121 = 0.02616 M.
:-)

Answer 2

Thanks Dr.Bob