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]

August 2, 2024

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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