If 4 moles of H 2 ​ are allowed to react with 1 mole of O 2 ​ , identify the limiting reagent and explain your answer with calculations.

📘 Limiting Reagent Calculation

Step 1: Write the balanced chemical equation

2H₂ + O₂ → 2H₂O

From the balanced equation:

2 mol H₂ : 1 mol O₂

This means 2 moles of H₂ require 1 mole of O₂.

Step 2: Given quantities

H₂ = 4 mol

O₂ = 1 mol

Step 3: Calculate O₂ required for 4 mol H₂

According to the equation:

2 mol H₂ → 1 mol O₂

Therefore, for 4 mol H₂:

O₂ required

= 4 × (1 / 2)

= 2 mol O₂

Step 4: Compare Required and Available O₂

O₂ Required = 2 mol
O₂ Available = 1 mol

Only 1 mole of O₂ is available, but 2 moles of O₂ are required to react completely with 4 moles of H₂.

❌ O₂ is insufficient.

🎬 Limiting Reagent Animation

The reaction requires: 2 H₂ : 1 O₂

H₂ H₂ → O₂ → H₂O H₂O

2H₂ + O₂ → 2H₂O

For 4 mol H₂, we need 2 mol O₂. But only 1 mol O₂ is available.

Step 5: How much H₂ actually reacts?

From the equation:

1 mol O₂ → 2 mol H₂

Since only 1 mol O₂ is available:

1 mol O₂ → 2 mol H₂

Therefore, only 2 moles of H₂ can react.

H₂ Available = 4 mol
H₂ Reacted = 2 mol
H₂ Left = 2 mol

Step 6: Calculate H₂O formed

According to the balanced equation:

1 mol O₂ → 2 mol H₂O

Therefore:

1 mol O₂ → 2 mol H₂O

Thus, 2 moles of water are formed.

✅ Final Answer

Limiting Reagent = O₂

Because 4 mol H₂ require 2 mol O₂, but only 1 mol O₂ is available.

O₂ is completely consumed first.

Excess H₂ remaining = 2 mol

H₂O formed = 2 mol

🎯 Quick Method to Find Limiting Reagent

1. Write the balanced equation.

2. Compare the given amounts with the stoichiometric ratio.

3. Find which reactant gets completely consumed first.

4. That reactant is the Limiting Reagent.

Less than required → Limiting Reagent