
In the above reaction the Ethyl group of grignard reagent acts as a strong base.
As per the stoichiometric equation the number of moles of methanol is equal to the number of moles of ethane.
Volume of methanol=Volume of ethane=2.24ml
Hence, number of moles of ethane
n=22.42.24×10−3=10−4
Weight of the gas= Number of moles× Molecular mass
W=n×M
=10−4×30=3mg