Molar mass from ammonia volume matches ammonium sulphate Steps:

• Moles of NH₃ produced = 2.51 dm³ / 24 dm³ mol⁻¹ = 0.1046 mol
• Assuming 1 mol NH₃ per mol salt, moles of salt = 0.1046 mol
• Molar mass of salt = 13.6 g / 0.1046 mol ≈ 130 g mol⁻¹
• Ammonium sulphate (D) has M = 132 g mol⁻¹, closest match Why D is correct:
• Calculated molar mass (130 g mol⁻¹) aligns with given value for ammonium sulphate (132 g mol⁻¹) per the 1:1 NH₃ assumption Why the others are wrong:
• A: M = 96 g mol⁻¹ and 2 NH₃ per formula unit yield 6.8 dm³, excess ammonia
• B: M = 53.5 g mol⁻¹ yields 6.1 dm³, excess ammonia
• C: M = 80 g mol⁻¹ yields 4.08 dm³, excess ammonia

Final answer: D