Question #a465d

1 Answer
anor277
Jun 5, 2017

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#?

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