For N₂ + 3 H₂ → 2 NH₃: starting with 1 mol N₂ and 5 mol H₂, the limiting reagent is
- A
Both equally
- Bcheck_circle
N₂
- C
H₂
- D
NH₃
Explanation
Stoichiometric ratio is N₂:H₂ = 1:3. With 1 mol N₂ we'd consume 3 mol H₂, leaving 2 mol H₂ unreacted. With 5 mol H₂ we'd need 5/3 ≈ 1.67 mol N₂ but only have 1 mol. So N₂ runs out first — N₂ is the limiting reagent; H₂ is in excess. The correct answer is <strong>A</strong> (N₂).