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Question

On heating 4.9 gms of KClO3, the weight loss is found to be 0.384 gms. The % of KClO3 decomposed is ___


Solution

2KClO32KCl+3O2
The reduction in weight is due the escape of Oxygen from the container.
Hence the weight of O2 is 0.384 gm
moles of O2=wtmol wt=0.38432=0.012 moles
2 moles of KClO3 gives 3 moles of O2.
x  moles of KClO3 gives 0.012 moles
2x=30.012
x=0.012×23
x = 0.008 moles
Wt of KClO3 decomposed = mol wt × moles = 122.5 × 0.008 = 0.98 gm
% dissociation = decomposed wttotal wt=0.984.9×100=20%.

 

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