Question

pure oxygen is prepared by thermal decomposition kclo3 according equation kclo3(s)giveskcl(s)+3÷2o2​

Answer 1

Answer:

The reaction is:

KClO3 ----> KCl + 32O2 Or, 2KClO3 -----> 2KCl + 3O2

So, Moles of KClO3 = MassMolar Mass = 12.25 g123 g/mol = 0.099 mol

Therefore, Moles of Oxygen = 3 x 0.099 mol = 0.297 mol

Volume of Oxygen liberated = 22.7 mol/L x 0.297 mol = 6.74 L

Answer 2

Answer:

3.36L of pure oxygen is prepared by thermal decomposition KClO₃.

Explanation:

It decomposes at 480°C according to the reaction 2KClO₃ 2KCl + 3O₂, with a mass loss of 39% anticipated. As a result, KClO₃ melts at 355.6 degrees Celsius and decomposes at 480 degrees Celsius.

Decomposition of KClO₃ takes place as:

2KClO₃(s)---> 2KCl(s) + 3O₂(g)

2 mole of KClO₃ = 3 mole of O₂

∴ 3 mole of O₂ formed by 2 mole of KClO₃

molar mass of KClO₃ = 39+35.5+3(16)=122.5

∴ 245g KClO₃ produce= 22.4×3L of oxygen

12.25g KClO₃ will produce= (22.4×3/24.5)×12.25

                                      = 3.36 L of oxygen.

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