How many litre of methane would be produce at NTP when 119 gm of CH3MgBr is treated with 6.5 gm of HC \equiv CH?

Engineering

Chemistry

Chemical Properties of Alkyl Halide

Reactions of Organometallic Compound Grignard Reagents

Question

How many litre of methane would be produce at NTP when 119 gm of CH₃MgBr is treated with 6.5 gm of HC $\equiv$ CH?

2.8 litre

11.2 litre

5.6 litre

22.4 litre

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Acetelene has two acidic hydrogens
2CH₃MgBr + HC $\equiv$ CH → 2CH₄ + (MgBr – C $\equiv$ C – MgBr)

$\frac{119}{119} = 1 mole \underset{L . R .}{\frac{6 .5}{26} = \frac{1}{4} mole}$

Moles of CH₄ produced = $\frac{1}{4} \times 2 = \frac{1}{2} moles$
Volume of CH₄ at NTP = $\frac{1}{2} \times 22.4 = 11.2 litre$ Ans.