Question

Decomposition of a `97.1*g` mass of potassium chlorate, `KClO_3`, yields what mass of dioxygen?

Answer 1

We need a stoichiometrically balanced equaition. Approximately #62.4" g" of dioxygen gas is produced.

Explanation:

We start with a stoichiometrically balanced equation:

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

To work well, this decompostion reaction requires a small quantity of `MnO_2` to act as a catalyst. The stoichiometry is the same.

`"Moles of KCl"` `=` `(97.1*g)/(74.55*g*mol^-1)` `=` `1.30*mol`

Given the stoichiometry, `3/2` equiv of dioxygen are evolved per equiv potassium chlorate.

And thus mass of `O_2` `=` `3/2xx1.30*molxx32.00*g*mol^-1` `~=` `62.4*g` dioxygen gas.