Question

Calculate the mass of potassium chlorate required to liberate 6.72 of oxygen at STP.

[Assume: molar mass of potassium chlorate as 122.5 g / mol].

Answer 1

$2KCl{O}_3⟶2KCl+3O_2$

From the above equation,

Amount of $KCl{O}_3$ required to liberate $(22.4\times 3)L$ of oxygen $=2\times 122.5g$

Therefore,

Amount of $KCl{O}_3$ required to liberate $6.72L$ of oxygen $=\frac{2\times 122.5}{22.4\times 3}\times 6.72=24.5g$

Hence, $24.5g$ of $KCl{O}_3$ is required.