Calculate the mass of potassium chlorate required to liberate

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calculate the mass of potassium chlorate required to liberate 6.72 dm³ of oxygen at stp, molar mass of potassium chlorate is 122.5 g/ mol.

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Preparation of oxygen in the laboratory can be done by decomposing of potassium chlorate.

Potassium chlorate decomposes, on strong heating, to form potassium chloride and oxygen.

This decomposition of potassium chlorate can be achieved at lower temperatures when the catalyst manganese (IV) oxide is used. 

The decomposition reaction of potassium chlorate is as follows:
             
 
                            Heat
     2KClO₃   ---------------------------->    2KCl  +  3O₂

The mole ratio for the reaction is 2:2:1 meaning, two moles of KClO₃ yields 2 moles of KCl and 3 moles of O₂

Calculate the moles of oxygen from the 6.72 dm³ given

1 dm³ = 1 liter 
6.72 dm³ =  6.72 liters of oxygen

Using the ideal gas law, which states that 1 mole of an ideal gas occupies 22.4 liters by volume we can calculate the moles of oxygen as follows:

If 22.4 liters = 1 mole
Then 6.72 liters  = 1 × 6.72/ 22.4
                           = 0.3 moles

From the equation above we found the mole ratio for the equation to be 2:2:1

That means, for 0.3 moles of oxygen to be liberated, we will need 0.3 × 2 = 0.6 moles of KClO₃

Use the 0.6 moles of KClO₃ to find the mass required

Moles = mass / molar mass      and,
 
mass = molar mass × moles
molar mass of potassium chlorate is 122.5 g/mol (given) and moles is 0.6(calculated)
         

Therefore mass of KClO₃ required   = 122.5 g/mol x 0.6 moles
                                                         = 73.5 g

Therefore you need 73.5 g of potassium chlorate to liberate 6.72 dm³  of oxygen. 

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