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

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Answer 1 · 2016-10-07T18:32:27

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

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.

Decomposition of a 97.1*g97.1*g mass of potassium chlorate, KClO_3KClO_3, yields what mass of dioxygen?