How to find the volume of oxygen release by the decomposition if potassium chlorate? In the reaction:
     $$\ce {2KClO3 -&gt; 2KCl + 3O2}$$
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 $\ce {PV=nRT}$.