Limiting and Excess Reactant: PECTAA Class 11 Chemistry, Punjab Board

You have 10 slices of bread and 3 slices of cheese, and every sandwich needs 2 slices of bread and 1 of cheese. How many sandwiches? Not 5. The cheese runs out after 3. In chemistry the cheese is the limiting reactant, and the leftover bread is the excess.

Section 4.7 of Chapter 4 (Stoichiometry) in the PECTAA Chemistry 11 book covers limiting and excess reactants. This note is for FSc Part 1 (1st year, Class 11) students in Lahore, Faisalabad, Sargodha, Rawalpindi, Multan, Gujranwala and the other Punjab boards. Gujrat district comes under the Gujranwala board. PECTAA writes the books. Your own board sets your paper.

What to know

In many processes the reactants are not present in the proportions of the balanced equation. The book says a large amount of an inexpensive reactant is often supplied, for three reasons:

• to make sure all of the expensive reactant is converted to the desired product
• to produce the maximum amount of product
• to increase the rate of reaction

Limiting reactant. Oxygen in excess makes things burn faster. The fuel is consumed earlier. The reactant consumed earlier is the limiting reactant (or limiting reagent). It controls the amount of product. Once it is used up, the reaction stops and no more product forms. The maximum amount of product depends on the amount of the limiting reactant.

Excess reactant. A reactant that is present in a larger amount than the stoichiometry needs, and remains unreacted at the end, is the excess reagent (or excess reactant).

Strategy (section 4.7.1). To identify the limiting reactant:

1. Calculate the moles of each reactant from the amounts given.
2. Use the balanced equation to find the moles of product each reactant could give.
3. The reactant that gives the least product is the limiting reactant. The other is in excess.

Key points

• Book example of the idea. For 2H₂ + O₂ → 2H₂O, 2 moles of H₂ (4 g) need only 1 mole of O₂ (32 g). If you mix 1 mole of H₂ with 1 mole of O₂, all the H₂ reacts and ½ mole of O₂ is left over.
• Use moles, not grams. Compare moles through the equation, not the masses.
• Product from the limiting reactant only. Calculate the mass of product from the limiting reactant, never from the excess one.
• Excess left over = amount you started with − amount used.
• Concentration route. For solutions the book uses n = C × V (moles = concentration in mol dm⁻³ × volume in dm³).

Worked examples

Example 1 (book Sample Problem 4.18). 1.25 dm³ of 0.05 mol dm⁻³ Pb(NO₃)₂ is mixed with 2.00 dm³ of 0.025 mol dm⁻³ Na₂SO₄. Find the mass of PbSO₄ formed. The net ionic equation is Pb²⁺ + SO₄²⁻ → PbSO₄.

1. Moles of Pb²⁺ = 0.05 × 1.25 = 0.0625 mol.
2. Moles of SO₄²⁻ = 0.025 × 2.00 = 0.05 mol.
3. They react 1 : 1, so SO₄²⁻ (0.05 mol) is used up first. It is the limiting reactant.
4. Moles of PbSO₄ = 0.05 mol.
5. Mass = 0.05 mol × 303.3 g mol⁻¹ = 15.2 g.

Example 2 (book Sample Problem 4.19). 30 g of CH₄ burn with 50 g of O₂ to give CO₂ and water. (a) Find the mass of CO₂. (b) Find the mass of excess reagent left.

1. Balanced equation: CH₄ + 2O₂ → CO₂ + 2H₂O.
2. Moles: CH₄ = 30 ÷ 16 = 1.875 mol. O₂ = 50 ÷ 32 = 1.563 mol.
3. CO₂ from CH₄: 1 × 1.875 = 1.875 mol. CO₂ from O₂: 0.5 × 1.563 = 0.7815 mol.
4. O₂ gives less CO₂, so O₂ is the limiting reactant.
5. Mass of CO₂ = 0.7815 mol × 44 g mol⁻¹ ≈ 34.4 g.
6. CH₄ used = 0.5 × 1.563 = 0.7815 mol = 12.5 g. CH₄ left = 30 − 12.5 = 17.5 g.

Example 3 (from the book's H₂ and O₂ discussion). 1 mole of H₂ and 1 mole of O₂ react to form water. Which is limiting and how much is left?

1. 2H₂ + O₂ → 2H₂O, so 1 mole of H₂ needs ½ mole of O₂.
2. H₂ is the limiting reactant. It is all used up.
3. O₂ left = 1 − ½ = ½ mole.

Common mistakes in the exam

1. Picking the reactant with the smaller mass as limiting. Compare moles through the equation.
2. Ignoring the coefficients. In Example 2, one mole of CH₄ needs two moles of O₂.
3. Calculating the product from the excess reactant. The limiting reactant sets the product.
4. Forgetting units. Write mol, g mol⁻¹ and g.
5. Mixing up C and V units. Volume must be in dm³ for n = CV.
6. Using an unbalanced equation. Balance it first.

Exam technique

• Write the balanced equation, then the moles, then the limiting reactant in words.
• Use the book's three steps as your three headings. Examiners see the method.
• Round at the end, not in the middle. Keep one extra digit while you work.
• Check your board's notice for the sections in your paper.

Practice questions

1. Define limiting reactant. Answer: The reactant that is consumed earlier and controls the amount of product formed.
2. Name two reasons for supplying a reactant in excess. Answer: To convert all of an expensive reactant, to get the maximum product, or to increase the rate of reaction (any two).
3. In 2H₂ + O₂ → 2H₂O, 2 mol H₂ and 2 mol O₂ are mixed. Which is limiting? Answer: H₂. It needs only 1 mol O₂, so 1 mol of O₂ is left.
4. How many moles of PbSO₄ form from 0.0625 mol Pb²⁺ and 0.05 mol SO₄²⁻? Answer: 0.05 mol, because SO₄²⁻ is limiting.
5. What is the formula for moles in a solution? Answer: n = C × V, with C in mol dm⁻³ and V in dm³.

Ready to practise this? Continue in 1st Year Chemistry.