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4.9 gram ${KClO}_{3}$ on calcination suffers a loss in weight of 0.384 grams. The percentage of ${KClO}_{3}$ that got decomposed (Atomic mass of K=39, Cl = 35.5, O=16)

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Detailed Solution

$2 {KClO}_{3 (s)} \to 2 {KCl}_{(s)} + 3 O_{2 (g)}$

$2 \times 122 .5 g$ = $3 \times 32 g$

x 0.384g

$x = 0.98 g$ of ${KClO}_{3}$

$%$ decomposed = $\frac{0.98}{4.9} \times 100 = 20$

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