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(A) The reaction is: $C_6H_5COOH + Br_2 \xrightarrow{FeBr_3} C_6H_4(Br)COOH + HBr$Molar mass of benzoic acid $(C_7H_6O_2)$ $= (7 	imes 12) + (6 	ime

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$78$

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(A) The reaction is: $C_6H_5COOH + Br_2 \xrightarrow{FeBr_3} C_6H_4(Br)COOH + HBr$Molar mass of benzoic acid $(C_7H_6O_2)$ $= (7 	imes 12) + (6 	imes 1) + (2 	imes 16) = 84 + 6 + 32 = 122 \,g/mol$.Molar mass of $m$-bromobenzoic acid $(C_7H_5BrO_2)$ $= 122 - 1 + 80 = 201 \,g/mol$.Moles of benzoic acid used $= \frac{6.1 \,g}{122 \,g/mol} = 0.05 \,mol$.According to the stoichiometry,$1 \,mol$ of benzoic acid produces $1 \,mol$ of $m$-bromobenzoic acid.Theoretical yield of $m$-bromobenzoic acid $= 0.05 \,mol 	imes 201 \,g/mol = 10.05 \,g$.Actual yield $= 7.8 \,g$.Percentage yield $= \frac{	ext{Actual yield}}{	ext{Theoretical yield}} 	imes 100 = \frac{7.8}{10.05} 	imes 100 \approx 77.61 \,\%$.Rounding off to the nearest integer,we get $78 \,\%$.

Consider the reaction shown below where $6.1 \,g$ of benzoic acid is used to obtain $7.8 \,g$ of $m$-bromobenzoic acid. The percentage yield of the product is ....... .
(Round off to the Nearest integer) [Given : Atomic masses : $C = 12.0 \,u, H = 1.0 \,u, O = 16.0 \,u, Br = 80.0 \,u$]

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Consider the reaction shown below where $6.1 \,g$ of benzoic acid is used to obtain $7.8 \,g$ of $m$-bromobenzoic acid. The percentage yield of the product is ....... .(Round off to the Nearest integer) [Given : Atomic masses : $C = 12.0 \,u, H = 1.0 \,u, O = 16.0 \,u, Br = 80.0 \,u$]