The decomposition of potassium chlorate (#KClO_3#) is used as a source of oxygen in the laboratory. How much potassium chlorate is needed to produce 19.7 mol of oxygen?

1 Answer
Feb 25, 2017

Answer:

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

Explanation:

Normally, this rxn must be calalyzed with a little #Mn(IV)# salt, but we address the question with this stoichiometry.

We spit out a #19.7*mol# quantity of dioxygen gas. Given the stoichiometry there were #2/3xx19.7*mol# #"potassium chlorate"#.

This represents a mass of #2/3xx19.7*molxx122.55*g*mol^-1# of #"potassium chlorate"# #~~# #1.6*kg#.

This is not a reaction that you do at home.