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What volume of oxygen gas is released at STP if 10.0 g of potassium chlorate is decomposed? (The molar mass of KClO3 is 122.55 g/mol.)

Verified Answer :

Hagrid

Given:

10g of potassium chlorate
decomposed

STP

molar mass of KClO3 is 122.55 g/mol

 

Required:

Volume of oxygen gas

 

Solution:

The decomposition reaction is

2KClO3 → 2KCl + 3O2

                Moles of O2 = 10g KClO3 (1
mol KClO3/122.55 g/mol KClO3)(3 moles O2/2 moles KClO3) = 15 moles O2

                 Using ideal gas law: PV = nRT

               

                PV = nRT

                V =
nRT/P

                V =
(15 moles O2)(0.08206 L-atm/mol-K)(273K)/1 atm

                V =
336.04 L O2