Question

Calculate the mass of potassium chlorate required to liberate 6.72 dm^{​​​​​3} of oxygen at STP

Answer 1

The decomposition reaction of potassium chlorate is as follows:


Heat
2KClO₃ ----------> 2KCl+ 3O₂

The mole ratio for the reaction is 2:2:3 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 2:2:3

3mol of O2 produced from 2 mol of KClO3
1mol of O2 from 2/3mol KClO3.

That means, for 0.3 moles of oxygen to be liberated, we will need 0.3 × 2/3 = 0.2moles of KClO₃

Use the 0.2 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 and moles is 0.2(calculated)


Therefore ,
mass of KClO₃ required = 122.5 g/mol x 0.2moles
= 24.5g