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Q.

If potassium chlorate is 80% pure then 48 g of oxygen would be produced by the thermal decomposition of
$2 {KClO}_{3} \to 2 KCl + 3 O_{2}$

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a

20 g of $K C l O_{3}$

b

153.12 g of $K C l O_{3}$

c

120 g of $K C l O_{3}$

d

90 g of $K C l O_{3}$

answer is A.

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Detailed Solution

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

$2 \times 122.5 g \to 3 \times 32 g$

$? - - - - 48 g = 122.5 g$

$80 % K C l O_{3} = \frac{122.5 \times 100}{80} = 153.12 g$

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