Question

The weight of sample of 10 percentage kclo3 required to produce 48 g of oxygen is

Answer 1

The weight of sample of 10 percentage kclo3 required to produce 48 g of oxygen is  1225gm.

The reaction involving heating of KClO3 is as follows:

2 KClO3 → 2 KCl + 3 O2

Molecular mass of 2 KClO3 = 2(39 + 35.5 + 3*16) = 245gm.

Weight of O2 produced in this reaction = 3 moles = 3(16*2) = 96gm.

∴ From the reaction, we get to know that 96gm of oxygen is produced from 245gm KClO3

So, 48gm (as mentioned in question) is produced from [(245×48) ÷ 96] gm KClO3 = 122.5gm.

So, 100% pure KClO3 requires 122.5gm of it to produce 48gm O2.

But it is mentioned that KClO3 is 10% pure, so amount of KClO3 required now will be:

[(100÷10)×122.5]gm = 1225gm KClO3.

So, 1225 gm of 10% pure KClO3 is required to produce 48gm O2.