15.8 grams of potassium chlorate is decompossed and the gas is colected over water at 29 degrees celcius when water vapor pressure is 30 mmHg. Assuming 100% yeild of products, what is the total pressure in the vessel after the reaction in complete?

Answer 1

2KClO3 ==> 2KCl + 3O2

mols KClO3 = 15.8g/molar mass
Using the coefficients in the balanced equation, convert mols KClO3 to mols O2. That is mols KClO3 x (3 mols O2/2 mols KClO3) = ?
Then PV = nRT to convert mols KClO3 to pressure in atm. I would convert that to mm Hg(multiply n*760) and add the vapor pressure of the H2O for total P.

Answer 2

what do you use for the V in PV=nRT?

Answer 3

I don't know and I didn't noticed when I posted the response that it was not listed. You must have it or have some way to calculate it. Do you have any other information about the problem.

Answer 4

no

Answer 5

The problem is impossible to do. There must have been a grammatical error somewhere