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

Answer 1

Noah Aldrich

Approx. #8*L#

We require (i) the following chemical reaction to demonstrate how potassium chlorate decomposes:

#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.

Additionally, (ii) we require the molar amount 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#

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Answer 2

Axel Alvarado

To determine the volume of dioxygen gas from the decomposition of potassium chlorate at 27 °C, additional information such as the quantity of potassium chlorate and the balanced chemical equation for the reaction is needed.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.