What is the minimum quantity (in grams) of methyl iodide required for preparing $1$ mole of ethane by Wurtz reaction?

[Atomic weight of iodine $= 127$ ]

142

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568

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326

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284

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Solution

The correct option is D $284$ g
The reaction is $2 C H_{3} I + 2 N a \to C_{2} H_{6} + 2 N a I$

One mole of ethane requires two moles of methyl iodide.

The molecular weight of methyl iodide is $142$ grams per mole.

Thus, two moles of methyl iodide will weigh $284$ grams.

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Similar questions

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

2.5

(C_{2} H_{6})

C = 12

H = 1

2 H I (g) ⇋ H_{2} (g) + I_{2} (g)

4.5

10

3

H_{2} (g) + I_{2} (g) \Leftrightarrow 2 H I (g)

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