Question

What volume of dioxygen gas would result from decomposition of potassium chlorate at `27` `""^@C`?

Answer 1

Approx. `8*L`

Explanation:

We need (i) a chemical reaction that illustrates the decomposition of potassium chlorate:

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

For this reaction to work well, generally a catalyst, typically `MnO_2(s)`, is added as an oxygen transfer reagent.

And (ii) we need the molar quantity of potassium chlorate:

`=` `(25*g)/(122.55*g*mol^-1)` `=` `0.204*mol`

Now clearly, a molar quantity of `0.204*molxx3/2` dioxygen gas are evolved, i.e. `0.306*mol`.

And thus, using the Ideal Gas equation: `V=(nRT)/P`

`=` `(0.306*molxx0.0821*L*atm*K^-1*mol^-1xx300*K)/((700*"Torr")/(760*"Torr"*atm^-1)`

`=` `??L`