How many grams of o2 can be prepared from 4.50g of kcio3

Take the molar mass of each element within the compound.

1 mole of the following...
K = 39g
Cl = 35g
O = 16g

Now find the molar mass of the compound itself.
K = 39
Cl = 35
O₃ = 16 * 3 = 48
39+35+48 = 122g

Set up a proportion with the amount of KClO₃ used, the molar mass of KClO₃, the amount of oxygen taken from a mole of KClO₃, and the amount of oxygen you will take from the amount of KClO₃ you have.



Cross multiply to get the following equation:

122x = 216

Divide both sides by 122 to solve for x.

x = 1.77

You can prepare 1.77g of O₂ from 4.5g of KClO₃.

Yeah, but what are you trying to ask?