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)#

Peyton Adams · Accepted Answer

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#

Consequently, given a stoichiometric reaction, we ought to obtain

#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#

Cooper Adams · Answer

undefined To find the volume of dioxygen gas produced when 2.25×10^24 formula units of KClO₃ decompose, you can use the stoichiometry of the reaction. The decomposition of KClO₃ produces KCl and O₂ gas.

First, calculate the number of moles of KClO₃:
1 formula unit of KClO₃ produces 1 molecule of O₂.
1 mole of O₂ gas at STP occupies 22.4 liters.
Convert formula units to moles:
$$ 2.25 \times 10^{24} \text{ formula units} \times \frac{1 \text{ mole}}{6.022 \times 10^{23} \text{ formula units}} = \text{moles of } KClO₃ $$

Then, use the stoichiometry of the reaction to find the moles of O₂ gas produced:
$$ \text{moles of } KClO₃ \times \frac{1 \text{ mole O₂}}{1 \text{ mole KClO₃}} = \text{moles of O₂ gas} $$

Finally, convert moles of O₂ gas to volume at STP:
$$ \text{moles of O₂ gas} \times 22.4 \text{ liters/mole} = \text{volume of O₂ gas at STP} $$

Plug in the values to find the volume of dioxygen gas produced.

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

Potassium chlorate undergoes the following thermal decomposition: #KClO_3(s) + Delta rarrKCl(s) + 3/2O_2(g)#

Potassium chlorate undergoes the following thermal decomposition:

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