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Weight of pure potassium chlorate required to decompose and produce 6 moles of oxygen is
(Atomic mass of K = 39; Atomic mass of Cl = 35.5)

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245 g

122.5 g

490 g

73.5 g

answer is B.

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Detailed Solution

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

$2 \times 122.5 g$ 3 moles

$W_{K C l O_{3}} = 245 \times 2 g$

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