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Weight of 50 ...

Question

Weight of 50% pure KClO3 required to produce 33.6 litres of O2 at STP is ?

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Solution

3 vols. of oxygen requires KClO₃ = 2 vols.

So, 1 vols. of oxygen will require KClO₃ =

So, 33.6 litres of oxygen will require KClO₃ = 2/3 * 33.6 =22.4Litre

22.4 litres of KClO₃ has mass = 1mole=122.5 g

Thereofore weight of KClO3 required to produce 33.6 litres of O2 = 122.5g

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