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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 $1~\mathrm{L} = 1~\mathrm{kg}$.

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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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  • $\begingroup$ You should guide the student to the answer, not provide it on a silver platter. Otherwise, the student doesn't learn. $\endgroup$ – LDC3 Dec 26 '14 at 14:50
  • $\begingroup$ I totally agree with you. But here the student must know this fundamental property of an ideal gas under NTP conditions. Apparently, he doesn't know it and it is frequently used in solving chemistry problems. So, I allowed for myself to solve the problem this time, as a role model problem. $\endgroup$ – Yomen Atassi Dec 26 '14 at 15:03
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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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  • $\begingroup$ @Martin You know, you could have added that yourself. $\endgroup$ – LDC3 Dec 26 '14 at 6:08
  • $\begingroup$ That's right, thought a comment would be appropriate to do so. $\endgroup$ – Martin - マーチン♦ Dec 26 '14 at 6:10
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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 $0.45 Γ— 24 = 10.8dm^3$. $1dm^3 = 1L$ so your answer would be $10.8L$

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