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
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⤋