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, $V=(nRT)/P$

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