What volume of dioxygen gas would be generated by complete decomposition of a 10*g mass of KClO_3, under a pressure of 750*mm*Hg?

Sep 3, 2017

Well, we need a stoichiometric equation......and I get under $2 \cdot L$.

Explanation:

$K C l {O}_{3} \stackrel{M n {O}_{2}}{\rightarrow} \frac{3}{2} {O}_{2} \left(g\right) + K C l$

Some $M n \left(I V\right)$ salt is usually added to catalyze the reaction.

$\text{Moles of potassium chlorate} = \frac{10 \cdot g}{122.55 \cdot g \cdot m o {l}^{-} 1}$

$= 0.0816 \cdot m o l$

And thus we gets $\frac{3}{2} \times 0.0816 \cdot m o l = 0.122 \cdot m o l$ $\text{dioxygen gas}$.

And then we simply solve the Ideal Gas equation.....

$V = \frac{n R T}{P} = \frac{0.0816 \cdot m o l \times 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 291 \cdot K}{\frac{750 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1}}$

=??*L

Note the pressure measurement. A mercury column is a VERY convenient means to measure gas pressure in that we know that $1 \cdot a t m$ will support a column of mercury that is $760 \cdot m m$ high. These days mercury has all but disappeared from laboratories because of perceived safety concerns. (I keep my manometers under lock and key, away from nosy safety inspectors).