Question
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?
Answers
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.
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.
what do you use for the V in PV=nRT?
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.
no
The problem is impossible to do. There must have been a grammatical error somewhere
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