From 6.55 g of aniline, the maximum amount of acetanilide that can be prepared will be ________ × 10⁻¹ g.

Solution

Chemistry ma’am · 4 steps

One aniline gives one acetanilide, so find the moles of aniline and convert them to grams of product.

1. The reaction. Acetylation swaps one H on the -NH₂ group for an acetyl group. One mole of aniline gives one mole of acetanilide.

   C₆H₅NH₂ + (CH₃CO)₂O → C₆H₅NHCOCH₃ + CH₃COOH

2. Molar masses. Aniline, C₆H₇N: 72 + 7 + 14 = 93 g mol⁻¹. Acetanilide, C₈H₉NO: 96 + 9 + 14 + 16 = 135 g mol⁻¹.
3. Put in numbers. Moles of aniline = 6.55/93 = 0.0704 mol. Mass of acetanilide = (6.55 × 135)/93 = 9.508 g.
4. So. 9.508 g = 95.08 × 10⁻¹ g, so the answer is 95.

The common slip: 9.5. Entering the mass in plain grams gives 9.5, not 95. The blank is in units of 10⁻¹ g, so multiply by 10.

The idea to keep. For a maximum yield question, write the balanced equation, read off the mole ratio, and carry moles from reactant to product. Then match the power of ten the question asks for.

Answer from the NTA final answer key. Solution by PrepWiser.