Question

Calculate the mass of potassium chlorate required to liberate $6.72$ $d{m}^3$ of oxygen at STP.

[Molar mass of potassium chlorate is $122.5$ g/mol].

Answer 1

Decomposition of potassium chlorate yields potassium chloride and oxygen as:

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

Thus 2 moles of potassium chlorate yields 3 moles of oxygen gas.

2 moles of potassium chlorate $=2\times 122.5=245g$ of potassium chlorate

At STP, the volume occupied by 1 mol of gas $=22.4d{m}^3$

the volume occupied by three moles of a gas $=3\times 22.4=67.2d{m}^3$

Therefore, 245g of potassium chlorate yields $67.2d{m}^3$ of oxygen gas

To liberate $6.72d{m}^3$ oxygen amount of potassium chlorate required is

$=\frac{245}{67.2}\times 6.72=24.5g$

Hence, $24.5g$ of potassium chlorate required to liberate $6.72d{m}^3$ of oxygen at STP