Question

Question #a465d

Answer 1

Approx. `3.5*g`

Explanation:

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

`KClO_3 stackrel(MnO_2,Delta)rarrKCl(s) + 3/2O_2(g)`

Note that the reaction is catalyzed by a bit of `Mn(IV)` salt. Heating without the catalyst would result in incomplete reduction to `KClO`.

And (ii) we need equivalent quantities of the reagents. Given `0.215*mol` 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*atm`, and `298*K`?