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

Oct 7, 2016

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

#### Explanation:

$K C l {O}_{3} \left(s\right) + \Delta \rightarrow K C l \left(s\right) + \frac{3}{2} {O}_{2} \left(g\right) \uparrow$
To work well, this decompostion reaction requires a small quantity of $M n {O}_{2}$ to act as a catalyst. The stoichiometry is the same.
$\text{Moles of KCl}$ $=$ $\frac{97.1 \cdot g}{74.55 \cdot g \cdot m o {l}^{-} 1}$ $=$ $1.30 \cdot m o l$
Given the stoichiometry, $\frac{3}{2}$ equiv of dioxygen are evolved per equiv potassium chlorate.
And thus mass of ${O}_{2}$ $=$ $\frac{3}{2} \times 1.30 \cdot m o l \times 32.00 \cdot g \cdot m o {l}^{-} 1$ $\cong$ $62.4 \cdot g$ dioxygen gas.