# Question a465d

Jun 5, 2017

Approx. $3.5 \cdot g$

#### Explanation:

We need (i), a stoichiometric equation......

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

Note that the reaction is catalyzed by a bit of $M n \left(I V\right)$ salt. Heating without the catalyst would result in incomplete reduction to $K C l O$.

And (ii) we need equivalent quantities of the reagents. Given $0.215 \cdot m o l$ of chlorate we should generate 3/2 equivs of dioxygen gas..............

3/2xx0.215*molxx32.00*g*mol^-1=??*g#.

This is a very convenient lab synthesis of dioxygen gas. How many litres of dioxygen would you get under standard conditions of $1 \cdot a t m$, and $298 \cdot K$?