A $90 \mathrm{g}$ of ethyl amine on reaction with methyl chloride produced a tertiary amine as an exclusive product. The amount of methyl chloride required is
[Given mass in amu : $\mathrm{H}=1, \mathrm{C}=12, \mathrm{N}=14, \mathrm{Cl}=35.5$]

The Standard Reaction is:

${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{NH}}_2 + 2{\mathrm{CH}}_3\mathrm{Cl} \to {\left({\mathrm{CH}}_3\right)}_3\mathrm{N} + \mathrm{HCl}$

Molar mass of ethyl amine $= 45 \mathrm{g}$

The molar mass of methyl chloride $= 50.5 \mathrm{g}$

Given Moles ethyl amine$= \frac{90}{45}=2$ 

As per the Standard reaction,

$1$ mole of ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{NH}}_2$ react with $2$ mole of ${\mathrm{CH}}_3\mathrm{Cl}$, 

So, $2$ moles of ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{NH}}_2$ react with $4$ moles of ${\mathrm{CH}}_3\mathrm{Cl}$

Hence, the amount methyl chloride obtained is $4\times 50.5 = 202 \mathrm{g}$