9.3 g of pure aniline is treated with bromin | vedclass

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(B) The reaction of aniline with bromine water is: $C_6H_5NH_2 + 3Br_2 \rightarrow C_6H_2Br_3NH_2 + 3HBr$ (Aniline) $\rightarrow$ ($2,4,6-$tribromoani

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(B) The reaction of aniline with bromine water is: $C_6H_5NH_2 + 3Br_2 ightarrow C_6H_2Br_3NH_2 + 3HBr$ (Aniline) $ightarrow$ ($2,4,6-$tribromoaniline) Molar mass of aniline $(C_6H_7N)$ = $93 \ g/mol$. Molar mass of $2,4,6-$tribromoaniline $(C_6H_4Br_3N)$ = $12 	imes 6 + 1 	imes 4 + 80 	imes 3 + 14 = 72 + 4 + 240 + 14 = 330 \ g/mol$. According to the stoichiometry,$93 \ g$ of aniline produces $330 \ g$ of $2,4,6-$tribromoaniline. Therefore,$9.3 \ g$ of aniline should theoretically produce: $\frac{330}{93} 	imes 9.3 = 33 \ g$ of $2,4,6-$tribromoaniline. The actual mass obtained is $26.4 \ g$. Percentage yield = $\frac{	ext{Actual yield}}{	ext{Theoretical yield}} 	imes 100$ Percentage yield = $\frac{26.4}{33} 	imes 100 = 80 \ \%$.

$9.3 \ g$ of pure aniline is treated with bromine water at room temperature to give a white precipitate of the product '$P$'. The mass of product '$P$' obtained is $26.4 \ g$. The percentage yield is $......... \ \%$.

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