Potassium perchlorate decomposes as follows: KClO4 (s) --> KCl (s) + 2 O2 (g)
If 269 mL of wet O2 (g) is collected over water at 25 c (vapor pressure= 23.8 torr) and the barometric pressure is 748.0 torr, how many grams of potassium perchlorate have decomposed?
3 answers
Use Ideal Gas Law (PV=nRT & n=PV/RT)=> moles of O2... Divide by 2 => moles of KClO3 ... Multiply moles KClO3 by formula wt of KClO3 => grams of KClO3 decomposed. Be sure all data is converted to same units as R-value. I get 0.021 gms KClO3. Good luck.
Correction... I'm used to decomposing KClO3, should of used f.Wt. of KClO4 = 138g/mole... Same logic applies to all else. Moles KClO4 decomposed = 0.0017mole O2, then => I get 0.235g KClO4 decomposed ... Sorry bout that.
correction-2 => 0.0017 mole KClO4! not O2.