Question

How much kclo3 should decompose to produce 67.2 l o2?

Answer 1

Hi.
 Reaction for the given question is-----
                                      2KClO3 = KCl + KClO+O2↑
Note-----This reaction needs to be heated in contact with a catalyst, typically manganese (IV) dioxide (MnO2).

From the equation,
                1 volume of the oxygen is produced by the 2 volume of KClO3
Therefore, 67.2 liter-------------------------------------------------  67.2 * 2 liter KClO3
                                                                                          =   134.4 liter.

Thus volume of potassium chlorate = 134.4 liter.

Hope it will help u.

If u have any query, u can ask me.

Please mark this answer as brainliest.

Answer 2

Answer:

134.4 liter.

Explanation:

Reaction for the given question is-----

                                     2KClO3 = KCl + KClO+O2↑

Note-----This reaction needs to be heated in contact with a catalyst, typically manganese (IV) dioxide (MnO2).

From the equation,

               1 volume of the oxygen is produced by the 2 volume of KClO3

Therefore, 67.2 liter-------------------------------------------------  67.2 * 2 liter KClO3

                                                                                         =   134.4 liter.

Thus volume of potassium chlorate = 134.4 liter.