Consider the reaction shown below. The percen | vedclass

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(C) The reaction is: $C_6H_5COCl + (C_6H_5)_2NH \rightarrow C_6H_5CON(C_6H_5)_2 + HCl$Molar mass of $C_6H_5COCl = 77 + 12 + 16 + 35.5 = 140.5 \, g/mol

Vedclass · Accepted Answer

$77$

Vedclass · Answer

(C) The reaction is: $C_6H_5COCl + (C_6H_5)_2NH ightarrow C_6H_5CON(C_6H_5)_2 + HCl$Molar mass of $C_6H_5COCl = 77 + 12 + 16 + 35.5 = 140.5 \, g/mol$.Molar mass of $C_6H_5CON(C_6H_5)_2 = 77 + 12 + 16 + 14 + 2(77) = 273 \, g/mol$.Moles of $C_6H_5COCl = \frac{0.140 \, g}{140.5 \, g/mol} \approx 0.001 \, mol$.Since the stoichiometry is $1:1$,theoretical moles of product $= 0.001 \, mol$.Theoretical mass of product $= 0.001 \, mol 	imes 273 \, g/mol = 0.273 \, g$.Actual mass of product formed $= 0.210 \, g$.Percentage yield $= \frac{	ext{Actual yield}}{	ext{Theoretical yield}} 	imes 100 = \frac{0.210}{0.273} 	imes 100 \approx 76.92 \%$.Rounding off to the nearest integer,we get $77 \%$.

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