Given #2.25xx10^24# formula units of #KClO_3#, what volume of dioxygen gas under standard conditions would result on heating?

Question

Potassium chlorate undergoes the following thermal decomposition:

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

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Answer 1 · 2017-05-09T10:04:32

You have the stoichiometric equation........and I get a volume of over #"80 litres.........."#

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

#"Moles of potassium chlorate"=(2.25xx10^24*"formula units")/(6.022xx10^23*"formula units"*mol^-1)=3.74*mol#

And thus, given stoichiometric reaction, we should get,

#3.74*molxx3/2*mol# #"dioxygen gas"# evolved, i.e. #5.60*mol#

And we use the Ideal Gas equation, #V=(nRT)/P#

#=(5.60*molxx0.0821*(L*atm)/(K*mol)xx556*K)/(3.00*atm)=??L#