Question

Calculate the mass of potassium chlorate required to liberate $6.72d{m}^3$ of oxygen at STP. Molar mass of potassium chlorate is $122.5g/mol$.

Answer 1

Heat

$2KCl{O}_3\stackrel{Heat}{\to }2KCl+3O_2$

The mole ratio for the reaction is 2:2:1 meaning, two moles of $KCl{O}_3$ yields 2 moles of KCl and 3 moles of $O_2$

Calculate the moles of oxygen from the $6.72d{m}^3$ given

$1d{m}^3=1liter$

$6.72d{m}^3=6.72liters$ 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 $KCl{O}_3$

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 $KCl{O}_3$ required = 122.5 g/mol x 0.6 moles

= 73.5 g

Therefore you need 73.5 g of potassium chlorate to liberate $6.72d{m}^3$ of oxygen.