9.3 g of aniline is subjected to reaction with excess of acetic anhydride to prepare acetanilide. The mass of acetanilide produced if the reaction is 100% completed is ______ g.

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
9.3 g of aniline is subjected to reaction with excess of acetic anhydride to prepare acetanilide. The mass of acetanilide produced if the reaction is 100% completed is ______ $\times 10^{- 1}$ g.

(Given molar mass in g ${mol}^{- 1}$ N : 14, O : 16
C : 12, H : 1) [2024]

Ans.
(135)

Moles of aniline $(n_{C_{6} H_{5} {NH}_{2}})$ $= \frac{Mass of aniline}{M_{C_{6} H_{5} {NH}_{2}}} = \frac{9.3 g}{93 g {mol}^{- 1}} = 0.1 mol$

As aniline is the limiting reagent, moles of acetanilide produced depend upon moles of aniline.

By stoichiometry:

moles of acetanilide = moles of aniline

moles of acetanilide $(n_{C_{6} H_{5} {NHCOCH}_{3}}) = 0.1 mol$

$Mass of acetanilide = n_{C_{6} H_{5} {NHCOCH}_{3}} \times M_{C_{6} H_{5} {NHCOCH}_{3}}$

$= 0.1 mol \times 135 g {mol}^{- 1}$

$= 13.5 g = 135 \times 10^{- 1} g$

Want Us to Email you About Special Offers & Updates?