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
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.
To determine the percent of O3 that reacts, you first need to calculate the moles of O3 that react.
From the balanced equation, it can be seen that the reaction consumes equal moles of NO and O3, with a 1:1 mole ratio. Therefore, since 2.50 moles of NO react, 2.50 moles of O3 will also react.
Now, you need to find the fraction of O3 that reacts out of its initial moles. To do this, divide the moles of O3 that react (2.50) by the initial moles of O3 (3.30):
(2.50 moles O3 / 3.30 moles O3) = 0.75757575
Next, convert this decimal to a percentage by multiplying by 100:
0.75757575 * 100 = 75.76%
So, the correct percent of O3 that reacts is approximately 75.76%, not 75.8% as you initially calculated.
In general, when determining the percent of a substance that reacts in a chemical reaction, you can use the following formula:
Percent Reacted = (moles reacted / initial moles) * 100
Just make sure to accurately determine the moles reacted and the initial moles to obtain the correct result.