Limiting reagent: Class 11 practice

What you should know first

• Stoichiometry

The idea behind limiting reagent

The limiting reagent runs out first and decides how much product forms. Compare the moles available with the ratio in the balanced equation.

Step-by-step method

1. Convert each reactant to moles.
2. Divide by its coefficient.
3. The smaller value is the limiting reagent.
4. Use it to find the product.

Worked example 1 (Basic)

8 g of H₂ reacts with 224 g of O₂ (2H₂ + O₂ → 2H₂O). Find the limiting reagent and the mass of water formed.

1. n(H₂) = 4, n(O₂) = 7
2. H₂ needs 2 mol O₂ → limiting: H₂
3. Water = 4 mol = 72 g
4. Answer: H₂ is limiting; 72 g water

Worked example 2 (Standard)

12 g of H₂ reacts with 160 g of O₂ (2H₂ + O₂ → 2H₂O). Find the limiting reagent and the mass of water formed.

1. n(H₂) = 6, n(O₂) = 5
2. H₂ needs 3 mol O₂ → limiting: H₂
3. Water = 6 mol = 108 g
4. Answer: H₂ is limiting; 108 g water

Worked example 3 (Challenge)

4 g of H₂ reacts with 32 g of O₂ (2H₂ + O₂ → 2H₂O). Find the limiting reagent and the mass of water formed.

1. n(H₂) = 2, n(O₂) = 1
2. H₂ needs 1 mol O₂ → limiting: neither (exact ratio)
3. Water = 2 mol = 36 g
4. Answer: neither (exact ratio) is limiting; 36 g water

Common mistakes and misconceptions

• Picking the reactant with less mass
• Ignoring coefficients

Practice questions

1. (Basic) 16 g of H₂ reacts with 160 g of O₂ (2H₂ + O₂ → 2H₂O). Find the limiting reagent and the mass of water formed.
2. (Basic) 12 g of H₂ reacts with 224 g of O₂ (2H₂ + O₂ → 2H₂O). Find the limiting reagent and the mass of water formed.
3. (Standard) 6 g of H₂ reacts with 160 g of O₂ (2H₂ + O₂ → 2H₂O). Find the limiting reagent and the mass of water formed.
4. (Standard) 10 g of H₂ reacts with 160 g of O₂ (2H₂ + O₂ → 2H₂O). Find the limiting reagent and the mass of water formed.
5. (Challenge) 14 g of H₂ reacts with 64 g of O₂ (2H₂ + O₂ → 2H₂O). Find the limiting reagent and the mass of water formed.
6. (Challenge) 12 g of H₂ reacts with 128 g of O₂ (2H₂ + O₂ → 2H₂O). Find the limiting reagent and the mass of water formed.

Answer key with full working

• Q1: n(H₂) = 8, n(O₂) = 5 → H₂ needs 4 mol O₂ → limiting: H₂ → Water = 8 mol = 144 g → Answer: H₂ is limiting; 144 g water
• Q2: n(H₂) = 6, n(O₂) = 7 → H₂ needs 3 mol O₂ → limiting: H₂ → Water = 6 mol = 108 g → Answer: H₂ is limiting; 108 g water
• Q3: n(H₂) = 3, n(O₂) = 5 → H₂ needs 1.5 mol O₂ → limiting: H₂ → Water = 3 mol = 54 g → Answer: H₂ is limiting; 54 g water
• Q4: n(H₂) = 5, n(O₂) = 5 → H₂ needs 2.5 mol O₂ → limiting: H₂ → Water = 5 mol = 90 g → Answer: H₂ is limiting; 90 g water
• Q5: n(H₂) = 7, n(O₂) = 2 → H₂ needs 3.5 mol O₂ → limiting: O₂ → Water = 4 mol = 72 g → Answer: O₂ is limiting; 72 g water
• Q6: n(H₂) = 6, n(O₂) = 4 → H₂ needs 3 mol O₂ → limiting: H₂ → Water = 6 mol = 108 g → Answer: H₂ is limiting; 108 g water

Syllabus note

This skill is listed under "Some Basic Concepts of Chemistry" for Class 11 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

Related practice and notes

• Some Basic Concepts of Chemistry chapter notes (Class 11)
• Molarity of solutions
• Mass–mass stoichiometry
• Empirical and molecular formula
• Theoretical and percentage yield
• Formula from combustion data
• Quantum numbers and orbitals
• Hydrogen spectrum and Rydberg formula
• Hybridisation and molecular shape (VSEPR)
• Hess's law and enthalpy of reaction
• All Class 11 practice

Frequently asked questions

Which chapter is limiting reagent in?

"Some Basic Concepts of Chemistry", Class 11 Chemistry (NCERT/CBSE). Other boards may place it differently.

How were the answers checked?

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.