# How many grams of KCIO_3 must be heated to produce 6.8 moles of oxygen?

Dec 4, 2016

Over $500 \cdot g$ $K C l {O}_{3} \left(s\right)$ are required; I assume you mean $\text{potassium chlorate}$.

#### Explanation:

We assess the decomposition reaction of $\text{potassium chlorate}$:

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

This reaction is typically catalyzed by $M n {O}_{2}$, which acts as an oxygen transfer agent.

And thus each equiv dioxygen gas requires $\frac{2}{3}$ equiv of $\text{potassium chlorate}$. Per equiv $\text{potassium chlorate}$, how much dioxygen do we get?

We require $6.8 \cdot m o l$ ${O}_{2} \left(g\right)$ and thus we need $\frac{2}{3} \times 6.8 \cdot m o l$ $\text{potassium chlorate}$.

i.e. $\frac{2}{3} \times 6.8 \cdot m o l \times 122.55 \cdot g \cdot m o {l}^{-} 1 \cong 0.6 \cdot k g$