posted by Anne
I don't understand part b
NO(g) + O3(g) <=> NO2(g) + O2(g)
(a) If 2.50 moles of NO and 3.30 moles of O3 are placed in a 5.30 L flask, what is the equilibrium concentration of O2? Kc = 5.00 ✕ 10-6.
(I got this correct)
(b) What percent of the O3 reacts?
My Work for B:
NO to O3 ratio is 1:1
2.50 moles of NO are used, 3.30 of O3
This means NO is limiting
1:1 ratio = 2.50:2.50 ratio
If O3 reacted 2.50 of its 3.30 moles, then 2.5/3.3 would give %
2.5/3.3 = .75757575 = 75.8%
This is incorrect
I wonder if you are trying to report too many significant figures? You are allowed only two; therefore, that 75.76 should be rounded to 76%. I assume are reporting to a data base. Often too man s.f. is the problem with these problems.
Unfortunately, 76% was also not the right answer
Well, I looked at your numbers and not the problem. You ARE allowed three s.f. I think your answer of 75.8% is correct. Some round to 75.7%. Other than that I don't know.