Question

Find the mass of kclo3, required to prepare 3.2g of o2.

Answer 1

Hi.

The equation for the above question will be-----
                         2KClO3 = KCl + KClO+O2↑
Note---This reaction needs to be heated in contact with a catalyst, typically manganese (IV) dioxide (MnO2)
   Molecular Mass of KClO3 = 123 g
    Molecular Mass of O2 = 32 g
Now, from the equation,

      Since,         32 g of the O2 is prepared from the 123 g of KClO3
Therefore,         1 g-----------------------------------------------123/32 g of KClO3
 Therefore,        3.2 g-----------------------------------------------(123/32) * 3.2
                                                                                         =  12.3 g
Thus,mass of KClO3 required to obtained 3.2 grams of oxgen is 12.3 grams.

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Answer 2

Gram molecular mass of potassium = 39 g and that of chlorine = 35 g and oxygen = 16 g
Here 
Mass of  KClO3 → 39 + 35 + 48 = 122 g 
Mass of oxygen in KClO3 = 48 g

Therefore
48 g of oxygen is prepared from =  122 g of KClO3

1 g of oxygen is prepared from = 122/48 g of KClO3

3.2 g of oxygen is prepared from = 122/48 * 3.2 of KClO3
                                                         = 8.1333 g of KClO3

Therefore
8.1333 g of KClO3 is required to prepare 3.2 of oxygen.