plllllllease help, very important for me?

Question

We never did anything even close to this in class, and i cant find help in the textbook, im desperate, please help me...

"Small quantities of oxygen gas are sometimes generated by heating KClO3 in the presence of MnO2 as a catalyst:



2KClO3(s) → 2KCl(s) + 3O2(g)



What volume (ml) of O2 is collected over water at 20.0 oC by the reaction of 0.261 g of KClO3 if the barometric pressure is 722.7 mm Hg
The vapor pressure of water at 20.0 oC is 17.5 mm Hg. "

Someone told me that this is the correct was to do the problem, however the answer is incorrect, can someone please help me?

2 KClO3---> KCl + 3O2

the molecular weight of KClO3 is 122.5

so you can see that 2*122.5 = 245g of KClO3 corresponds to 3 moles of O2

0.261 g corresponds to 0.261*3/122.5 = 0.064 moles of O2

to calculate the volume you use the formula pV = nRT ;
T = 293 K and 1mmHg = 133 Pa

V = 0.064*8.31 * 293/ 722.7*133 =0.0016m^3 = 1.6 liter

Answer 1

A possible reason this is incorrect is when you plugged your numbers into PV=nRT, you only had 1 mmHg while the problem reads 17.5 mmHg (2333.14 Pa). Also, you are to subtract this from the barometric pressure to get your actual pressure, not multiply (according to the link mei gave). I don't remember doing problems exactly like this, but it appears otherwise you have set it up correctly. The only other thing I can suggest is to double check your units on R. You didn't list them.

Answer 2

http://answers.yahoo.com/question/index;_ylt=AuzEC1.QkBjRLN3bfEhrB0UjzKIX?qid=20061113014355AANWKiV

use that site as a guideline