Limiting Reagent – IGCSE Chemistry Definition and Key Facts

The limiting reagent concept is tested in Paper 4 of the Cambridge 0620 syllabus, often combined with mole calculations, percentage yield, and excess reagent questions. Identifying the limiting reagent is a Supplement skill that separates competent calculators from those who guess.

The 0620 definition

The limiting reagent (or limiting reactant) is the reactant that is completely consumed in a reaction, limiting the amount of product that can form. Once the limiting reagent runs out, the reaction stops — regardless of how much of the other reactant remains.

Step-by-step method to identify the limiting reagent

1. Write the balanced equation
2. Calculate the moles of each reactant given (using n = m/M or n = c x V)
3. Divide each number of moles by the coefficient (the number in front) in the balanced equation
4. The reactant with the smaller result is the limiting reagent

Worked example

Question: 4.8 g of magnesium reacts with 100 cm³ of 2.0 mol/dm³ hydrochloric acid. Mg + 2HCl → MgCl₂ + H₂. Identify the limiting reagent.

Step 1: Balanced equation — Mg + 2HCl → MgCl₂ + H₂

Step 2: Calculate moles

• Moles of Mg = 4.8 / 24 = 0.20 mol
• Moles of HCl = 2.0 x 0.100 = 0.20 mol

Step 3: Divide by coefficients

• Mg: 0.20 / 1 = 0.20
• HCl: 0.20 / 2 = 0.10

Step 4: HCl gives the smaller value (0.10 < 0.20), so HCl is the limiting reagent.

The maximum product is calculated from the limiting reagent: 0.20 mol HCl produces 0.10 mol H₂ (from the 2:1 ratio).

Why the limiting reagent matters

1. Determines product amount — the mass of product formed depends only on the limiting reagent, not the excess
2. Practical chemistry — in salt preparation, an excess of insoluble reactant is added to ensure all the acid (limiting) reacts, then the excess solid is filtered off
3. Industrial processes — one reactant is often made limiting deliberately; in the Haber process, the ratio of N₂ to H₂ is controlled, and unreacted excess is recycled

Linking to percentage yield

Once you know the maximum amount of product from the limiting reagent (the theoretical yield), you can calculate percentage yield:

Percentage yield = (actual yield / theoretical yield) x 100

Common scenarios in exam questions

Scenario	Approach
Given masses of two reactants	Convert both to moles; compare using coefficients
Given mass of one and volume + concentration of the other	Use n = m/M for the solid, n = c x V for the solution
Asked “which reactant is in excess?”	Find the limiting reagent first; the other is in excess
Asked to calculate the mass of product	Use moles of limiting reagent and the molar ratio

Second worked exam question

4.0 g of iron reacts with 4.0 g of sulfur: Fe + S → FeS. Determine which reactant is the limiting reagent and calculate the mass of FeS produced. (Ar: Fe = 56, S = 32) [4]

Moles of Fe = 4.0 / 56 = 0.0714 mol [1]. Moles of S = 4.0 / 32 = 0.125 mol [1]. Ratio is 1:1 in the equation. Fe has fewer moles, so Fe is the limiting reagent [1]. Mass of FeS = 0.0714 x (56 + 32) = 0.0714 x 88 = 6.3 g [1].

Common exam mistakes

1. Comparing raw moles without dividing by coefficients. If the equation shows 2HCl, you must divide the moles of HCl by 2 before comparing.
2. Calculating product from the excess reagent instead of the limiting reagent — the answer will be too large.
3. Assuming equal masses mean equal moles. 4 g of Fe and 4 g of S have different numbers of moles because they have different molar masses.
4. Forgetting to convert cm³ to dm³ when using c x V (divide by 1000).