Chem equilibrium

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.
.00121
(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

asked by Anne
  1. 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.

    posted by DrBob222
  2. Unfortunately, 76% was also not the right answer

    posted by Anne
  3. 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.

    posted by DrBob222

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