The percent loss in weight after heating a pure sample of potassium chlorate (mol.wt.=122.5) will be?
( ${K C l O}_{3} \to {K C l + O}_{2}$ )

12.25

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24.5

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39.18

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Solution

The correct option is A 39.18
We have,

$2 K C l O_{3} \to 2 K C l + 3 O_{2}$

molar mass of $K C l O_{3}$ = 122.5 g

mass of KCl = 74.5 g

so, percentage loss = $\frac{122.5 - 74.5}{122.5} * 100 = 39.18$

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Similar questions

w t . = 122.5

= 122.5

%

= 122.5

K C l O_{3} (s) \to K C l (s) + O_{2} (g)

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