Question

2.84 gms of methyl iodide was completely converted into $C{H}_{3}MgI$ and the product was decomposed by the excess of ethanol. The volume of the gaseous hydrocarbon produced at NTP will be_________.

• A. 22.4 lit.
• B. 2240 lit.
• C. 0.448 lit.
• D. 224 lit.

Answer 1

Answer: (C)

0.448 lit.

In this reaction, one mole of Grignard reagent reacts with one mole of alcohol to produce hydrocarbon.

$C{H}_{3}I\to C{H}_{3}MgI+C_2{H}_{5}OH\to C{H}_4+C_2{H}_{5}OMgI$

Mass of methyl iodide = 2.84 gm (given)
No. of mole of methyl iodide = 2.84/142 =0.02

So hydrocarbon produced = 0.02 mole

So the volume of produced hydrocarbon = 0.448 lit
(one mole produced 22.4 lit).

Option C is correct.