For the reaction 2H 2 ​ +O 2 ​ →2H 2 ​ O, 5 mol of H 2 ​ reacts with 2 mol of O 2 ​ . Identify the limiting reagent and calculate the mass of water formed.

📘 Complete Step-by-Step Solution

📌 Given Data

H₂ = 5 mol

O₂ = 2 mol

Reaction:
2H₂ + O₂ → 2H₂O

🔵 Step 1: Understand the Mole Ratio

From the balanced chemical equation:

2H₂ + O₂ → 2H₂O

2 mol H₂ : 1 mol O₂ : 2 mol H₂O

Therefore, 2 moles of H₂ require 1 mole of O₂.

🟢 Step 2: Find Which Reactant Is Required

For 5 mol H₂, the required amount of O₂ is:

O₂ required = 5 × (1/2)

= 2.5 mol O₂

⚠️ O₂ available = 2 mol

O₂ required = 2.5 mol

Therefore, O₂ is insufficient.

🔴 Step 3: Identify the Limiting Reagent

✅ O₂ is the Limiting Reagent.

Because the available 2 mol O₂ can react with only 4 mol H₂, while 5 mol H₂ is available.

2 mol O₂ × 2 mol H₂ / 1 mol O₂

= 4 mol H₂

Thus, 4 mol H₂ reacts and 1 mol H₂ remains unreacted.

⚛️ Reaction Visualization

::contentReference[oaicite:0]{index=0}

🟣 Step 4: Calculate Moles of Water Formed

From the equation:

1 mol O₂ → 2 mol H₂O

Therefore:

2 mol O₂ → ?

= 2 × 2

= 4 mol H₂O

✅ Water formed = 4 mol H₂O

🟠 Step 5: Calculate Molar Mass of H₂O

Molar Mass of H₂O

= (2 × 1) + 16

= 2 + 16

= 18 g mol⁻¹

🟢 Step 6: Calculate Mass of Water

Mass = Number of Moles × Molar Mass

= 4 × 18

= 72 g

💧 Mass of H₂O formed = 72 g

🎬 3D-Style Reaction Animation

The limiting reagent O₂ determines the maximum amount of water formed.

H H + O O

2H₂ + O₂ → 2H₂O

2 mol O₂ → 4 mol H₂O

4 mol H₂O → 72 g H₂O

⚡ Quick Shortcut

5 mol H₂ + 2 mol O₂
↓
O₂ required for 5 mol H₂ = 2.5 mol
↓
Only 2 mol O₂ available
↓
O₂ = Limiting Reagent
↓
2 mol O₂ → 4 mol H₂O
↓
4 × 18
↓
72 g H₂O

✅ Final Answer

Limiting Reagent = O₂

H₂O formed = 4 mol

Mass of H₂O = 72 g

🎯 Formula Revision

Balanced Equation → Mole Ratio
2H₂ : 1O₂ : 2H₂O
Limiting Reagent → O₂
2 mol O₂ → 2 mol H₂O
4 mol H₂O × 18 g mol⁻¹ = 72 g