For the reaction N2​+3H2​→2NH3​, 28 g of N2​ reacts with 6 g of H2​. Determine the limiting reagent, mass of NH3​ formed and mass of excess reagent left.

📘 Complete Step-by-Step Solution

📌 Given Data

Mass of N₂ = 28 g

Mass of H₂ = 6 g

Reaction: N₂ + 3H₂ → 2NH₃

🔵 Step 1: Calculate Molar Masses

Molar Mass of N₂

= 2 × 14

= 28 g mol⁻¹

Molar Mass of H₂

= 2 × 1

= 2 g mol⁻¹

Molar Mass of NH₃

= 14 + (3 × 1)

= 17 g mol⁻¹

🟢 Step 2: Calculate Moles of Reactants

Moles = Mass ÷ Molar Mass

N₂: 28 ÷ 28

= 1 mol

H₂: 6 ÷ 2

= 3 mol

N₂ = 1 mol    |    H₂ = 3 mol

🔴 Step 3: Identify the Limiting Reagent

From the balanced equation:

1 mol N₂ + 3 mol H₂ → 2 mol NH₃

The available quantities are exactly:

1 mol N₂ + 3 mol H₂

Required Ratio = Available Ratio

✅ Neither reactant is in excess.
Both N₂ and H₂ are completely consumed.

There is no excess reagent.

🎬 3D-Style Reaction Animation

N₂ + H₂ H₂ H₂ → NH₃ NH₃

1 N₂ + 3 H₂ → 2 NH₃

🟣 Step 4: Calculate Mass of NH₃ Formed

From the balanced equation:

1 mol N₂ → 2 mol NH₃

We have 1 mol N₂, therefore:

NH₃ formed = 2 mol

Molar mass of NH₃ = 17 g mol⁻¹

Mass = Moles × Molar Mass

= 2 × 17

= 34 g

✅ Mass of NH₃ formed = 34 g

🟠 Step 5: Calculate Mass of Excess Reagent Left

Since the reactants are present in the exact stoichiometric ratio:

1 mol N₂ requires exactly 3 mol H₂

Available H₂ = 3 mol

Required H₂ = 3 mol

Excess H₂ = Available H₂ − Required H₂

= 3 − 3

= 0 mol

Mass of excess H₂ = 0 × 2

= 0 g

✅ Excess reagent left = 0 g

Both N₂ and H₂ are completely consumed.

✅ Final Answer

Limiting Reagent = None

N₂ and H₂ are in exact stoichiometric ratio.

Mass of NH₃ formed = 34 g

Mass of excess reagent left = 0 g

🎯 Quick Revision

28 g N₂ = 1 mol N₂
6 g H₂ = 3 mol H₂
1 N₂ : 3 H₂ : 2 NH₃
2 mol NH₃ × 17 = 34 g NH₃
Excess reagent = 0 g