If potassium chlorate is 80% pure, then 48 gm of oxygen would be produced from (atomic mass of K =39) 153.12 gm of KClO3 122.5 gm of KClO3 245 gm of KClO3 98 gm of KClO3
Question
If potassium chlorate is 80% pure, then 48 gm of oxygen would be produced from (atomic mass of K =39) 153.12 gm of KClO3 122.5 gm of KClO3 245 gm of KClO3 98 gm of KClO3
Solution
The molecular weight of KClO3 is calculated as follows:
K = 39 Cl = 35.5 O3 = 16*3 = 48
So, the molecular weight of KClO3 = 39 + 35.5 + 48 = 122.5 g/mol
In KClO3, 3 atoms of oxygen are present. So, 48 g of oxygen is present in 122.5 g of KClO3.
But the given KClO3 is 80% pure. So, 100% pure KClO3 = 122.5 g / 80% = 153.12 g
Therefore, 48 g of oxygen would be produced from 153.12 g of KClO3.
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