Decomposition of a #97.1*g# mass of potassium chlorate, #KClO_3#, yields what mass of dioxygen?

1 Answer
Oct 7, 2016

Answer:

We need a stoichiometrically balanced equaition. Approximately #62.4" g" of dioxygen gas is produced.

Explanation:

We start with a stoichiometrically balanced equation:

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

To work well, this decompostion reaction requires a small quantity of #MnO_2# to act as a catalyst. The stoichiometry is the same.

#"Moles of KCl"# #=# #(97.1*g)/(74.55*g*mol^-1)# #=# #1.30*mol#

Given the stoichiometry, #3/2# equiv of dioxygen are evolved per equiv potassium chlorate.

And thus mass of #O_2# #=# #3/2xx1.30*molxx32.00*g*mol^-1# #~=# #62.4*g# dioxygen gas.