How to find the volume of oxygen release by the decomposition if potassium chlorate?

In the reaction:
 $$\ce {2KClO3 -> 2KCl + 3O2}$$
What is the volume of oxygen released under NTP conditions when $36.75~\mathrm{g}$ of $\ce {KClO3}$ is heated?


I tried to find the answer this way:


* $2~\mathrm{mol}$ $\ce{KClO3}$ form $3~\mathrm{mol}$ $\ce{O2}$
* $0.3~\mathrm{mol}$ $\ce{KClO3}$ form $0.45~\mathrm{mol}$ $\ce {O2}$, i.e. $36.75~\mathrm{g}~\ce{KClO3} = 0.3~\mathrm{mol}~\ce{KClO3}$


I don't know how to convert moles into litre. Can we convert mass directly to volume in $\mathrm{L}$. I read somewhere that $1~\mathrm{L} = 1~\mathrm{kg}$.




1 kg for 1 liter is, in fact, liquid water at room temperature. 


You need to use the ideal gas law $\ce {PV=nRT}$.


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In fact, under NTP conditions, each mole of an ideal gas occupies the volume $22.4\ce{L }$.


In your case, the volume of $0.45\ce{mol }$ of $\ce{O\_2 }$ is $22.4\times 0.45=10.08\ce{L }$.


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Avogadro's law states that "*equal volumes of all gases, at the same temperature and pressure, have the same number of molecules*", ie at constant temperature and pressure, 1 mole of any gas will occupy the same volume. At NTP, this volume is ~$24dm^3$. Therefore the volume of $0.45mol$ of $O\_2$ will be $0.45 × 24 = 10.8dm^3$. $1dm^3 = 1L$ so your answer would be $10.8L$


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