Calculate the volume of oxygen at STP that can be produced by 12.25g of KClO3?

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Solution

KClO3 Decomposes to give KCl and O2
2KClO3 → 2KCl +3O2

Molecular weight of KClO3 is 39 +35.3+48 = 122.5 grams

Here 2 moles of KClO3 has taken, Hence molecular weight = 2X122.5 = 245 grams

According to the equation 245g of KClO3 Produces 3 moles of Oxygen
i.e 3X22.4 L = 67.2 L of oxygen is produced
At STP 1 mole equals 22.4 litres
245g of KClO3 → 67.2 L of oxygen
12.25 g of KClO3 → ? oxygen
= 12.25 X 67.2/245
=3.36 L
Volume of Oxygen released = 3.36 L required answer.

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