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When 6.32g of potassium chlorate decomposes, determine the volume of oxygen produced at 105kpa and 300k

Accepted Answer

At the given conditions, 1 mole of any gas occupies 24.1dm3.

The decomposition reaction of potassium chlorate is:

2KClO3(s) --→ 2KCl(s) + 3O2(g)

No of moles of KClO3= given mass/molar mass= 6.32g/122.55g/mol=0.0515mol.

Therefore, moles of O2 produced= 3/2 * moles of KClO3= 3/2 * 0.0515mol=0.0773mol.

Volume of O2 produced at 105kPa and 300K = moles * molar volume =0.0773mol * 24.1dm3/mol = 1.87dm3

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