What is the minimum quantity of methyl iodide required for preparing one mole of ethane by Wurtz reaction ?(At.wt.of iodine=127)

142 gram

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568 gram

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326 gram

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284 gram

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Solution

The correct option is D 284 gram
In Wurtz reaction, two moles of alkyl halide reacts with sodium metal to form one mole of higher alkane.

Given reaction,
$2 C H_{3} I + 2 N a \to C H_{3} - C H_{3} + 2 N a I$

Molecular weight of methyl iodide $= 127 + 15 = 142$ gm

Hence, the minimum quantity of methyl iodide required $= 2 \times 142 = 284$ gm.

Hence, option $D$ is correct.

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= 127

The minimum weight in grams of methyl iodide required for preparing one mole of ethane by wutrz reaction.

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