how many moles of O2 gas will be obtained by heating 24.5 g potassium chlorate.
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he formula for the decomposition reaction of potassium chlorate is
2KClO3→2KCl+3O2
That gives you 3 moles of oxygen for each 2 moles of potassium perchlorate.
The molecular weight of potassium perchlorate is 122.55 g/mol.
24.5 g is roughly 24.5122.55=0.2 mol
The number of mole of oxygen is therefore 0.3 mol.
At STP the volume can be approximated by using the ideal gas equation.
V=nRTP=0.3×8.314×273101.3≈6.8 litres
Plz refer to the link for the answers
https://www.quora.com/What-is-the-volume-of-oxygen-which-can-be-obtained-from-24-5-g-of-potassium-chlorate-at-STP
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