How many litres of methane would be produced when $0.595$ gm of $C H_{3} M g B r$ is treated with excess of $C_{4} H_{9} N H_{2}$ ?

0.8 litre

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0.08 litre

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0.112 litre

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1.12 litre

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Solution

The correct option is D 0.112 litre
One mole of Grignard reagent will give one mole of methane.
Mole of Grignard reagent used $= \frac{0.595}{119} = 0.005$ moles
So, mole of methane produced $= 0.005$ moles
So, volume of methane produced $= 0.005 \times 22.4 = 0.112$ litre

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