Question

The decomposition of potassium chlorate (`KClO_3`) is used as a source of oxygen in the laboratory. How much potassium chlorate is needed to produce 19.7 mol of oxygen?

Answer 1

`KClO_3(s) + Delta rarr KCl(s) + 3/2O_2(g)`

Explanation:

Normally, this rxn must be calalyzed with a little `Mn(IV)` salt, but we address the question with this stoichiometry.

We spit out a `19.7*mol` quantity of dioxygen gas. Given the stoichiometry there were `2/3xx19.7*mol` `"potassium chlorate"`.

This represents a mass of `2/3xx19.7*molxx122.55*g*mol^-1` of `"potassium chlorate"` `~~` `1.6*kg`.

This is not a reaction that you do at home.