50% of the reagent is used for dehydrohalogen | vedclass

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(B) The chemical reaction for the dehydrohalogenation of ethyl chloride is: $CH_3CH_2Cl + KOH \xrightarrow{	ext{Ethanol}, \Delta} CH_2=CH_2 + KCl + H

Vedclass · Accepted Answer

$1.4$

Vedclass · Answer

(B) The chemical reaction for the dehydrohalogenation of ethyl chloride is: $CH_3CH_2Cl + KOH \xrightarrow{	ext{Ethanol}, \Delta} CH_2=CH_2 + KCl + H_2O$ Molar mass of $CH_3CH_2Cl = (2 	imes 12) + (5 	imes 1) + 35.5 = 64.5 \ g \ mol^{-1}$. Molar mass of the main product,ethene $(CH_2=CH_2)$ $= (2 	imes 12) + (4 	imes 1) = 28 \ g \ mol^{-1}$. According to the stoichiometry,$64.5 \ g$ of $CH_3CH_2Cl$ produces $28 \ g$ of ethene. Therefore,$6.45 \ g$ of $CH_3CH_2Cl$ would theoretically produce: $	ext{Mass} = \frac{6.45 \ g}{64.5 \ g \ mol^{-1}} 	imes 28 \ g \ mol^{-1} = 2.8 \ g$ of ethene. Since only $50\%$ of the reagent is used,the actual amount of product obtained is: $2.8 \ g 	imes 0.5 = 1.4 \ g$.

$50\%$ of the reagent is used for dehydrohalogenation of $6.45 \ g$ $CH_3CH_2Cl$. What will be the weight of the main product obtained (in $g$)? [At. mass of $H$,$C$,and $Cl$ are $1$,$12$,and $35.5 \ g \ mol^{-1}$ respectively]

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