A 10.0 g sample of ethane (C₂H₆) reacts with excess chlorine to give 1‑chloroethane (CH₃CH₂Cl) and 2‑chloroethane (CH₃CHCl₂) in a 2:1 ratio. Calculate the mass of 1‑chloroethane produced.

medium 2 marks Class 11 · Chemistry · Organic Chemistry Some Basic Principles and Techniques

Answer

14.3 g

The correct answer is 14.3 g — Finally, mass of 1‑chloroethane = 0.2219 mol × 64.51 g mol⁻¹ = 14.3 g (rounded to one decimal place).

Step-by-step solution

Calculate the molar mass of ethane: 2×12.01 + 6×1.008 = 30.07 g mol⁻¹.
Find moles of ethane: 10.0 g ÷ 30.07 g mol⁻¹ = 0.3328 mol.
The two products are formed in a 2:1 molar ratio, so total moles of products = 0.3328 mol. Thus moles of 1‑chloroethane = (2/3)×0.3328 = 0.2219 mol.
Compute molar mass of 1‑chloroethane: 2×12.01 + 5×1.008 + 35.45 = 64.51 g mol⁻¹.
Finally, mass of 1‑chloroethane = 0.2219 mol × 64.51 g mol⁻¹ = 14.3 g (rounded to one decimal place).

Common mistake: Using the molar mass of chlorine instead of the product in the final calculation

Concept tested: Stoichiometry of halogenation