When 32.25 g of ethyl chloride undergoes deh | vedclass

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(D) The chemical reaction for the dehydrohalogenation of ethyl chloride $(C_2H_5Cl)$ is: $C_2H_5Cl \rightarrow C_2H_4 + HCl$ Molar mass of $C_2H_5Cl =

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$7$

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(D) The chemical reaction for the dehydrohalogenation of ethyl chloride $(C_2H_5Cl)$ is: $C_2H_5Cl ightarrow C_2H_4 + HCl$ Molar mass of $C_2H_5Cl = (2 	imes 12) + (5 	imes 1) + 35.5 = 64.5 \ g/mol$. Molar mass of ethene $(C_2H_4)$ = $(2 	imes 12) + (4 	imes 1) = 28 \ g/mol$. According to the stoichiometry,$64.5 \ g$ of $C_2H_5Cl$ produces $28 \ g$ of $C_2H_4$. Therefore,$32.25 \ g$ of $C_2H_5Cl$ would theoretically produce: $(32.25 / 64.5) 	imes 28 = 0.5 	imes 28 = 14 \ g$ of $C_2H_4$. Since the yield is $50\%$,the actual mass of the product obtained is: $14 \ g 	imes 0.50 = 7 \ g$.

When $32.25 \ g$ of ethyl chloride undergoes dehydrohalogenation,it yields $50\%$ of the theoretical alkene product. What is the mass of the product obtained in $g$? (Atomic mass: $Cl = 35.5, C = 12, H = 1$)

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