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‘x’ g of KClO3 on decomposition gives ‘y’ ml of O2 at STP. The % purity of KClO3 would be
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'A'. Can you explain this answer?
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‘x’ g of KClO3 on decomposition gives ‘y’ ml o...
The balanced eqn. will be:
2KClO3 = 2KCl+3O2
The no. of moles of Oxygen = y/22.4
3 moles of Oxygen is produced from 2 moles of KClO3
⇒ 1 mole will be produced from 2/3 moles of KClO3
∴ y/22.4 will be producing from (2/3) × (y/22.4)
Hence, the mass of KClO3 used = (2/3) × (y/22.4)×M
% Purity = (2 × y × M)/(3 × 22.4 × x)
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‘x’ g of KClO3 on decomposition gives ‘y’ ml o...
The balanced eqn. will be
2KClo3 = 2KCl+3O2
the no. of moles of Oxygen = y/22.4

3 moles of Oxygen if produced from 2moles of KClo3 ...then 1 mole will produce from 2/3moles of KClo3 ...And y/22.4 will be producing from 2/3×y/22.4

,..no. of moles of KClo3 = 2/3×y/22.4
So..mass of KClo3 used =2/3×y/22.4×M(mol. wt.)

%purity = 2/3×y/22.4×M divide the whole to x
%purity=2×y×M/3×22.4×x
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