In the above reaction,3.9 g of benzene on ni | vedclass

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(B) The chemical reaction for the nitration of benzene is: $C_6H_6 + HNO_3 \xrightarrow{H_2SO_4} C_6H_5NO_2 + H_2O$Molar mass of benzene $(C_6H_6)$ $=

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$80$

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(B) The chemical reaction for the nitration of benzene is: $C_6H_6 + HNO_3 \xrightarrow{H_2SO_4} C_6H_5NO_2 + H_2O$Molar mass of benzene $(C_6H_6)$ $= 6 	imes 12 + 6 	imes 1 = 78 \ g/mol$.Molar mass of nitrobenzene $(C_6H_5NO_2)$ $= 6 	imes 12 + 5 	imes 1 + 14 + 2 	imes 16 = 123 \ g/mol$.According to the stoichiometry,$1 \ mole$ of benzene produces $1 \ mole$ of nitrobenzene.So,$78 \ g$ of benzene should theoretically produce $123 \ g$ of nitrobenzene.Therefore,$3.9 \ g$ of benzene should produce: $\frac{123}{78} 	imes 3.9 = 6.15 \ g$ of nitrobenzene.The actual amount of nitrobenzene formed is $4.92 \ g$.Percentage yield $= \frac{	ext{Actual yield}}{	ext{Theoretical yield}} 	imes 100 = \frac{4.92}{6.15} 	imes 100 = 80 \%$.

In the above reaction,$3.9 \ g$ of benzene on nitration gives $4.92 \ g$ of nitrobenzene. The percentage yield of nitrobenzene in the above reaction is............. $\%$. (Round off to the Nearest Integer).
(Given atomic mass: $C : 12.0 \ u$,$H : 1.0 \ u$,$O : 16.0 \ u$,$N : 14.0 \ u$)

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In the above reaction,$3.9 \ g$ of benzene on nitration gives $4.92 \ g$ of nitrobenzene. The percentage yield of nitrobenzene in the above reaction is............. $\%$. (Round off to the Nearest Integer).(Given atomic mass: $C : 12.0 \ u$,$H : 1.0 \ u$,$O : 16.0 \ u$,$N : 14.0 \ u$)