What is the volume of oxygen produced with 15.0 l of nitrogen in this decomposition reaction: KNO3 --> K2O(s) + N2(g) + O2(g)?

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When potassium nitrate KNO3 undergoes decomposition, the reaction is:

KNO3 --> K2O(s) + N2(g) + O2(g)

This has to be balanced, which gives:

4KNO3 --> 2K2O(s) + 2N2(g) + 5O2(g)

4 moles of KNO3 give 2 moles of nitrogen and 5 moles of oxygen. Here the moles of KNO3 that are reacting is not given, but it is given that the decomposition yields 15 l of nitrogen.

As the temperature and pressure for both nitrogen and oxygen produced is the same, we don't have to bother about the molar mass and the volume of a mole of each gas released is the same irrespective of their mass. The volume of oxygen produced would be 2.5 times the volume of nitrogen produced. As 2.5*15 = 37.5, 37.5 liters of oxygen is produced.

The volume of oxygen produced is 37.5 l

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