How many grams of #KCIO_3# must be heated to produce 6.8 moles of oxygen?

1 Answer
anor277
Dec 4, 2016

Over #500*g# #KClO_3(s)# are required; I assume you mean #"potassium chlorate"#.

Explanation:

We assess the decomposition reaction of #"potassium chlorate"#:

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

This reaction is typically catalyzed by #MnO_2#, which acts as an oxygen transfer agent.

And thus each equiv dioxygen gas requires #2/3# equiv of #"potassium chlorate"#. Per equiv #"potassium chlorate"#, how much dioxygen do we get?

We require #6.8*mol# #O_2(g)# and thus we need #2/3xx6.8*mol# #"potassium chlorate"#.

i.e. #2/3xx6.8*molxx122.55*g*mol^-1~=0.6*kg#

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