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

Oct 24, 2016

Approx. $8 \cdot L$

#### Explanation:

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

$K C l {O}_{3} \left(s\right) + \Delta \rightarrow K C l \left(s\right) + \frac{3}{2} {O}_{2} \left(g\right) \uparrow$

For this reaction to work well, generally a catalyst, typically $M n {O}_{2} \left(s\right)$, is added as an oxygen transfer reagent.

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

$=$ $\frac{25 \cdot g}{122.55 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.204 \cdot m o l$

Now clearly, a molar quantity of $0.204 \cdot m o l \times \frac{3}{2}$ dioxygen gas are evolved, i.e. $0.306 \cdot m o l$.

And thus, using the Ideal Gas equation: $V = \frac{n R T}{P}$

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

$=$ ??L