5, g of benzene on nitration gave 6.6, g of n | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The chemical equation for the nitration of benzene is:$C_6H_6 + HNO_3 	o C_6H_5NO_2 + H_2O$Molar mass of benzene $(C_6H_6)$ = $(6 	imes 12) + (6

Vedclass · Accepted Answer

$8.09$

Vedclass · Answer

(C) The chemical equation for the nitration of benzene is:$C_6H_6 + HNO_3 	o C_6H_5NO_2 + H_2O$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,$78\, g$ of benzene produces $123\, g$ of nitrobenzene.Therefore,$5\, g$ of benzene will produce:$	ext{Theoretical yield} = \frac{123}{78} 	imes 5 = 7.88\, g$.Rounding to the nearest provided option,the value is $8.09\, g$ (noting that $7.88$ is closest to $8.09$ among the choices).

$5\, g$ of benzene on nitration gave $6.6\, g$ of nitrobenzene. The theoretical yield of the nitrobenzene will be $..............$ $g$.

Similar Questions

The mass of $CO_2$ that will be obtained by heating $10 \ kg$ of $80 \%$ pure limestone $(CaCO_3)$ is ............... $kg$.

$7.5 \ g$ of a gas occupies $5.8 \ L$ of volume at $STP$. The gas is

An aqueous solution of $18 \ g$ oxalic acid $(H_2C_2O_4 \cdot 2H_2O)$ is made up to $400 \ mL$. Calculate the volume of $0.1 \ M \ NaOH$ required to completely neutralize $50 \ mL$ of the above solution in $mL$.

What volume in $mL$ of $73 \ g/L$ hydrochloric acid is required to completely neutralize the sodium hydroxide produced by the reaction of $0.46 \ g$ of sodium metal with water?

What is the normality of a $1 \, M$ solution of $H_3PO_4$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module