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 KClO₃ = 1.25 g

MOlar mass of KClO₃ = 122.55 g/mol

Moles of KClO₃ = Mass of KClO₃ / MOlar mass of KClO₃ =1.25 g/ 122.55 g/mol = 0.0102 moles

Reaction is:

2KClO₃(s) 2KCl (s) + 3O₂ (g)

FRom reaction, 2 mol of KClO₃ produces 3 mol of O₂

Thus,  0.0102 moles of KClO₃ produces 3*0.0102 / 2 = 0.0153 mol (n)of O₂

Total pressure = 751 torr

Pressure of water at 30 ^oC = 32 torr

Pressure of O₂ (P)=Total pressure -Pressure of water at 30 ^oC

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

temperature (T) =30 ^oC = 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