Acetanilide from Aniline | JEE Mains PYQ | Mole Concept
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 ______ × 10⁻¹ g.

(Given molar mass in g mol⁻¹ : N = 14, O = 16, C = 12, H = 1)

Explanation
Step 1: Write reaction stoichiometry

Aniline + Acetic anhydride → Acetanilide + Acetic acid
Mole ratio of aniline : acetanilide = 1 : 1

Step 2: Calculate molar masses

Molar mass of aniline (C₆H₇N) = (6×12) + (7×1) + 14 = 93 g mol⁻¹
Molar mass of acetanilide (C₈H₉NO) = (8×12) + (9×1) + 14 + 16 = 135 g mol⁻¹

Step 3: Calculate moles of aniline

Mass of aniline = 9.3 g
Moles of aniline = 9.3 ÷ 93 = 0.1 mol

Step 4: Calculate mass of acetanilide formed

Moles of acetanilide formed = 0.1 mol
Mass = 0.1 × 135 = 13.5 g

Step 5: Express in required form

13.5 g = 135 × 10⁻¹ g

Correct Answer = 135

Updated for JEE Main 2026: This PYQ is important for JEE Mains, JEE Advanced and other competitive exams. Practice more questions from this chapter.

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