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Question

What volume of oxygen at s.t.p. will be obtained by the action of heat on 20 g KClO3 [K = 39, Cl = 35.5, O = 16]

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Answer

2KClO3   +    Δ   →   KCl     +    3O2

245 g                                         67.2 lit

245 g of KClO3 gives 67.2 lit of O2

∴ 20 g of KClO3 will give = 67.2\over 245 × 20

= 5.486 lit of O2

Hence, 5.486 lit of oxygen will be obtained

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