90 g of ethylamine on reaction with methyl c | vedclass

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(C) The reaction for the formation of a tertiary amine from ethylamine $(CH_3CH_2NH_2)$ and methyl chloride $(CH_3Cl)$ is: $CH_3CH_2NH_2 + 2CH_3Cl \ri

Vedclass · Accepted Answer

$202$

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(C) The reaction for the formation of a tertiary amine from ethylamine $(CH_3CH_2NH_2)$ and methyl chloride $(CH_3Cl)$ is: $CH_3CH_2NH_2 + 2CH_3Cl ightarrow CH_3CH_2N(CH_3)_2 + 2HCl$ Molar mass of ethylamine $(C_2H_7N)$ $= (2 	imes 12) + (7 	imes 1) + 14 = 45 \ g/mol$. Molar mass of methyl chloride $(CH_3Cl)$ $= 12 + (3 	imes 1) + 35.5 = 50.5 \ g/mol$. Moles of ethylamine $= \frac{90 \ g}{45 \ g/mol} = 2 \ mol$. According to the stoichiometry of the reaction,$1 \ mol$ of ethylamine reacts with $2 \ mol$ of methyl chloride. Therefore,$2 \ mol$ of ethylamine will react with $2 	imes 2 = 4 \ mol$ of methyl chloride. Mass of methyl chloride required $= 4 \ mol 	imes 50.5 \ g/mol = 202 \ g$.

$90 \ g$ of ethylamine on reaction with methyl chloride produces a tertiary amine as the exclusive product. The amount of methyl chloride required is:
$[\text{Given atomic masses in amu}: H=1, C=12, N=14, Cl=35.5]$ (in $g$)

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$90 \ g$ of ethylamine on reaction with methyl chloride produces a tertiary amine as the exclusive product. The amount of methyl chloride required is: $[\text{Given atomic masses in amu}: H=1, C=12, N=14, Cl=35.5]$ (in $g$)