Question

How much potassium chlorate will be required to get 0.1 mole of
oxygen? (K=39, CI=35.5, O=16)​

Answer 1

Answer:

Oxygen is prepared by catalytic; decomposition of potassium chlorate (KClO

3

). Decomposition, of potassium chlorate gives potassium chloride (KCl) and oxygen (O

2

). How many moles and how many grams of KClO

3

are required to produce 2.4 mole O

2

?

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Decomposition of KClO

3

takes place as:

2KClO

3

(s)→2KCl(s)+3O

2

(g)

2 mole of KClO

3

=3 mole of O

2

∵ 3 mole O

2

formed by 2 mole KClO

3

∴2.4 mole of O

2

will be formed by:

(

3

2

×2.4) mol KClO

3

= 1.6 mole of KClO

3

.

Mass of KClO

3

= Number of moles × Molar mass

=1.6×122.5=196g

Answer 2

Explanation:

answer.

9.5gm of potassium chloride will be required to get 0.1 mol