Question

2.84 g of methyl iodide was completely converted into methyl magnesium iodide and the product was decomposed by an excess of ethanol.
The volume of the gaseous hydrocarbon produced at NTP will be

• A. 22.4 liter
• B. 22400 ml
• C. 0.488 liter
• D. 0.244 liter

Answer 1

Answer: (C)

0.488 liter

Moles of methyl iodide in 2.84 g of methyl iodide = $\frac{2.84}{142}$ = 0.02 moles

Reaction: $C{H}_{3}I+Mg\stackrel{DryEther}{\to }C{H}_{3}MgI$

So, the moles of methyl magnesium iodide produced are 0.02 moles.

Next Reaction: $C{H}_{3}MgI+ROH\to C{H}_4+\left(RO\right)MgI$

So, the moles of hydrocarbon formed are 0.02 moles.

At STP, 1 mole of gas occupies 22.4 L of volume.

$⟹$ 0.02 mole of methane will occupy 0.488 L of volume.

Hence, Option "C" is the correct answer.