What volume of dioxygen gas would result from decomposition of potassium chlorate at #27# #""^@C#?

1 Answer
Oct 24, 2016

Answer:

Approx. #8*L#

Explanation:

We need (i) a chemical reaction that illustrates the decomposition of potassium chlorate:

#KClO_3(s) + Delta rarr KCl(s) + 3/2O_2(g)uarr#

For this reaction to work well, generally a catalyst, typically #MnO_2(s)#, is added as an oxygen transfer reagent.

And (ii) we need the molar quantity of potassium chlorate:

#=# #(25*g)/(122.55*g*mol^-1)# #=# #0.204*mol#

Now clearly, a molar quantity of #0.204*molxx3/2# dioxygen gas are evolved, i.e. #0.306*mol#.

And thus, using the Ideal Gas equation: #V=(nRT)/P#

#=# #(0.306*molxx0.0821*L*atm*K^-1*mol^-1xx300*K)/((700*"Torr")/(760*"Torr"*atm^-1)#

#=# #??L#