Question

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.)

Answer 1

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