# chemistry

. A 0.428-gram sample of a mixture of the solids, KCl and KClO3, is heated. If 80.7 mL of oxygen gas is collected over water at 18.0 oC and 100.8 kPa, calculate the mass percentage of KClO3 in the mixture. (PH2O at 18.0 oC = 15.5 torr

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1. I assume that 18.0 is 18mm Hg.
2KClO3 ==> 2KCl + 3O2
Calculate mols O2 from PV = nRT
Since the pressure is listed in kPa and the vapor pressure in mm Hg you must convert one of them. I would convert 18.0 mm to kPa.
18.0 mm Hg x (1 atm/760 mm Hg) x (101.325 kPa/1 atm) = 2.399 kPa.
Ptotal = 100.8 = pO2 + pH2O
pH2O = 2.399. Solve for pO2 and substitute that for P in PV = nRT
You know V(in L), R(8.314) and T (in kelvin).
mols O2 x (2 mol KClO3/3 mols O2) = mols KClO3.
g KClO3 = mols KClO3 x molar mass KClO3
%KClO3 = (g KClO3/mass sample)*100 = ?

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