1.86 g of aniline completely reacts to form acetanilide. 10% of the product is lost during purificat — Amines Chemistry Question

💡 Solution & Explanation

**Step 1: Calculate molar masses** - Aniline (C₆H₅NH₂): 93 g/mol - Acetanilide (C₆H₅NHCOCH₃): 135 g/mol **Step 2: Calculate moles of aniline** Moles of aniline = 1.86 g ÷ 93 g/mol = 0.02 mol **Step 3: Use stoichiometry (1:1 molar ratio)** The reaction is: C₆H₅NH₂ + CH₃COCl → C₆H₅NHCOCH₃ + HCl Moles of acetanilide produced = 0.02 mol **Step 4: Calculate theoretical mass of acetanilide** Mass of acetanilide (theoretical) = 0.02 mol × 135 g/mol = 2.70 g **Step 5: Account for 10% loss during purification** If 10% is lost, 90% remains: Mass after purification = 2.70 g × 0.90 = 2.43 g **Step 6: Express in required format** 2.43 g = 243.00 × 10⁻² g Therefore, the answer is 243.00.