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Q.

Potassium chlorate $K C l O_{3}$ is used in fire works and explosives. It can be prepared by bubbling chlorine into hot aqueous potassium hydroxide; $K C l (a q)$ and $H_{2} O$ are the other products in the reaction. How many grams of $K C l O_{3}$ can be obtained from $138 L$ of $C l_{2}$ whose pressure is $760 m m H g$ at $27^{o} C$
$(A t w t s : K = 39,, O = 16, H = 1)$

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answer is 229.

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Detailed Solution

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$n = \frac{P V}{R T} = \frac{(760 / 760 a t m) (138 L)}{0.0821 \times 300} = 5.60 6 K O H + 3 C l_{2} \to K C l O_{3} + 5 K C l + 3 H_{2} O 3 m o l s o f C l_{2} g i v e 1 m o l o f K C l O_{3} ∴ M a s s o f K C l O_{3} = 5.60 \times \frac{1 m o l K C l O_{3}}{3 m o l C l_{2}} \times 122.5 = 228.78$

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