When 32.25 g of ethyl chloride is dehydrohal | vedclass

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(D) The reaction for dehydrohalogenation of ethyl chloride is: $C_2H_5Cl ightarrow C_2H_4 + HCl$Molar mass of $C_2H_5Cl = (2 	imes 12) + (5 	imes

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$7$

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(D) The reaction for dehydrohalogenation of ethyl chloride 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 $C_2H_4 = (2 	imes 12) + (4 	imes 1) = 28 \ g/mol$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: $\frac{28 	imes 32.25}{64.5} = 14 \ g$ 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 is dehydrohalogenated,it gives $50\%$ yield of alkene. What is the mass of the product in $g$? (Atomic mass of $Cl = 35.5$)

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