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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. "

2006-11-13 17:36:20 · 3 answers · asked by Paul S 2 in Science & Mathematics Chemistry

Where are these sililiar questions?...

2006-11-13 17:54:33 · update #1

Still dont know how to do it... sry, can you please explain in more detail?

2006-11-13 17:57:20 · update #2

I tried ur way and got the wrong method, right now im just checking if the numbers were calculated right, hopefully thats the problem

2006-11-13 18:20:26 · update #3

3 answers

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

2006-11-13 17:58:27 · answer #1 · answered by maussy 7 · 0 0

Just figure out the mass of oxygen that will be liberated.

Then, turn that mass into moles of oxygen.

Now, as you know that 1 mole is equal to 22.4 litres, you can use this relation to get the volume of oxygen liberated.

Best of luck.

2006-11-14 01:45:27 · answer #2 · answered by King of Hearts 6 · 0 0

some people have already posted up similar questions. just go search for similar questions

2006-11-14 01:52:17 · answer #3 · answered by mei l 1 · 0 0

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