Question

P. Calculate the mass of potassium
chlorate required to liberate 6.72 dm of
oxygen at STP. Molar mass of KCIO, is
122.5 g mol-1.​

Answer 1

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Calculate the mass of potassium chlorate required to liberate 6.72 dm

3

of oxygen at STP. Molar mass of potassium chlorate is 122.5 g/mol.

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ANSWER

Heat

2KClO

3

Heat

2KCl+3O

2

The mole ratio for the reaction is 2:2:1 meaning, two moles of KClO

3

yields 2 moles of KCl and 3 moles of O

2

Calculate the moles of oxygen from the 6.72 dm

3

given

1 dm

3

=1 liter

6.72 dm

3

=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 KClO

3

Use the 0.6 moles of KClO₃ to find the mass required

Moles = mass / molar mass

Then,

mass = molar mass × moles

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

Therefore mass of KClO

3

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 dm

3

of oxygen.

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