Calculate percentage atom economy when 46 g | vedclass

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Question

(B) The chemical reaction for the preparation of ethanol from chloroethane is: $C_2H_5Cl + KOH \rightarrow C_2H_5OH + KCl$. The molar mass of $C_2H_5C

Vedclass · Accepted Answer

$38.17$

Vedclass · Answer

(B) The chemical reaction for the preparation of ethanol from chloroethane is: $C_2H_5Cl + KOH ightarrow C_2H_5OH + KCl$. The molar mass of $C_2H_5Cl$ is $64.5 \ g/mol$,$KOH$ is $56 \ g/mol$,$C_2H_5OH$ is $46 \ g/mol$,and $KCl$ is $74.5 \ g/mol$. The formula for percentage atom economy is: $	ext{Atom Economy} = \frac{	ext{Molar mass of desired product}}{	ext{Total molar mass of all reactants}} 	imes 100$. Total molar mass of reactants = $64.5 + 56 = 120.5 \ g/mol$. Molar mass of desired product (ethanol) = $46 \ g/mol$. Atom Economy = $(46 / 120.5) 	imes 100 \approx 38.17 \%$. Thus,the correct option is $B$.

Calculate percentage atom economy when $46 \ g$ ethanol is obtained from $64.5 \ g$ chloroethane and $56 \ g$ $KOH_{(aq)}$. (in $\%$)

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