Limiting Reagent Problem

Limiting Reagent Problem

Example: Reaction Between Hydrogen and Oxygen to Form Water

Suppose you have 5.0 grams of hydrogen (H₂) and 32.0 grams of oxygen (O₂), and they react to form water (H₂O) according to the balanced chemical equation:

2H₂ + O₂  →  2H₂O

Step 1: Calculate the Moles of Each Reactant

First, we need to find the number of moles of hydrogen and oxygen:

• Molar mass of H₂ = 2.02 g/mol
• Molar mass of O₂ = 32.00 g/mol

Moles of H2 = 5.0 g / 2.02 g/mol ≈ 2.48 mol

Moles of O2 = 32.0 g / 32.00 g/mol = 1.00 mol

Step 2: Determine the Limiting Reagent

From the balanced equation, we see that 2 moles of H₂ react with 1 mole of O₂ to produce 2 moles of H₂O. Therefore, we can calculate the maximum amount of H₂O that can be produced by each reactant:

• Hydrogen can produce 2.48 mol H₂ / 2 mol H₂ × 2 mol H₂O =2.48 mol H₂O
• Oxygen can produce 1.00 mol O₂ × 2 mol H₂O = 2.00 mol H₂O

Since oxygen can produce fewer moles of H₂O, it is the limiting reagent.

Step 3: Calculate the Amount of Product Formed

The amount of water produced is limited by the amount of oxygen, the limiting reagent:

Moles of H₂O produced=2.00 mol

Step 4: Convert Moles of Product to Grams (if needed)

If required, you can convert the moles of water produced to grams using the molar mass of water (H₂O = 18.02 g/mol):

Mass of H₂O produced = 2.00 mol × 18.02 g/mol = 36.04 g 

Conclusion

In this reaction, oxygen is the limiting reagent, and the maximum amount of water that can be produced is 36.04 grams.