Question

Predict the volume of oxygen that would be produced by the decomposition of 1.25 g KClO3 according to the reaction equation:

2 KClO3 (s) 2 KCl (s) + 3 O2 (g)

The oxygen is collected over water at 30 oC at a total pressure of
751 torr; the vapor

pressure of water at 30 oC is 32 torr.

Answer #1

Mass of KClO_{3} = 1.25 g

MOlar mass of KClO_{3} = 122.55 g/mol

Moles of KClO_{3} = Mass of KClO_{3} / MOlar
mass of KClO_{3} =1.25 g/ 122.55 g/mol = 0.0102 moles

Reaction is:

2KClO_{3}(s)
2KCl (s) + 3O_{2} (g)

FRom reaction, 2 mol of KClO_{3} produces 3 mol of
O_{2}

Thus, 0.0102 moles of KClO_{3} produces
3*0.0102 / 2 = 0.0153 mol (n)of O_{2}

Total pressure = 751 torr

Pressure of water at 30 ^{o}C = 32 torr

Pressure of O_{2} (P)=Total pressure -Pressure of water
at 30 ^{o}C

= 751 torr-32 torr = 719 torr = 719/760 = 0.946 atm

temperature (T) =30 ^{o}C = 30+273 = 303K

Gas constant (R) = 0.08206 L.atm/mol.K

Ideal gas equation is:

PV= nRT

V = nRT/P = ( 0.0153 mol*0.08206 L.atm/mol.K*
303K)/ 0.946 atm = **0.4021
L**

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