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Limiting Reactant

Limiting reactant: identify what runs out first using moles and the balanced equation, then calculate theoretical yield and excess left.

  • SEC G3 Pure Chemistry 2027
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Learning objectives

  • calculate stoichiometric reacting masses and volumes of gases (one mole of gas occupies 24 dm3 at room temperature and pressure); calculations involving the idea of limiting reactants may be set (knowledge of the gas laws and the calculations of gaseous volumes at different temperatures and pressures are not required)

Limiting reactant questions become systematic when you use moles and the balanced equation. The limiting reactant is the one that runs out first, so it controls the maximum product.

1. Definition

A. Limiting Reactant

The limiting reactant is the reactant that is completely used up first, so it limits the amount of product formed.

B. Excess Reactant

The excess reactant is the reactant that is left over after the reaction stops.

C. Theoretical Yield

The theoretical yield is the maximum amount of product calculated from the limiting reactant (assuming the reaction goes to completion and no losses occur).

2. Key Ideas

A. Convert Everything to Moles First

Limiting reactant is about particle numbers, so you must work in moles, not grams or cm³.

B. Start From the Balanced Equation

The coefficients (the numbers in front of formulas) are the reacting mole ratios. If the equation is unbalanced, every answer after that is wrong.

C. Fast Test (Moles ÷ Coefficient)

For each reactant:

  1. Find moles.
  2. Divide by its coefficient in the balanced equation.

The smallest value is the limiting reactant.

D. RTP Reminder (For Gas Volumes)

At RTP (room temperature and pressure), the molar gas volume is 24 dm³/mol.

Recall: Moles, concentration, molar volume

If converting mass, solution concentration or gas volume into moles is still difficult, revise that step first: Mole & Molar Mass Molar Volume & Concentration Chemical Equations

3. Detailed Explanations

Quick Recall (limiting reactant test)
  • Convert every reactant to moles first (mass, solution, or gas).
  • Use the balanced equation: coefficients are the reacting mole ratios.
  • Quick test: compute “moles ÷ coefficient” for each reactant; the smallest is limiting.
  • For gases at RTP: n = V/24 with V in dm³.

A. Method 1: Moles ÷ Coefficient (Quick and Clean)

  1. Write the balanced equation.
  2. Calculate moles of every reactant.
  3. Compute “moles ÷ coefficient” for each reactant.
  4. The smallest value is limiting.

B. Method 2: “How Much Product Can Each Make?”

  1. Use each reactant (one at a time) to calculate the moles of product it could form.
  2. The reactant that gives the smaller product is the limiting reactant.

This method is slower but it is hard to mess up.

C. After You Find the Limiting Reactant

  • Use the limiting reactant to calculate moles (and mass/volume) of product.
  • If asked for the excess reactant left: calculate how much excess reactant was used, then subtract from the starting amount.

4. Common Mistakes

A. Stoichiometry Mistakes

  • Choosing the limiting reactant simply because it has the smaller mass; reactants must be compared in moles and against their coefficients.
  • Comparing moles but ignoring coefficients (also wrong).
  • Using Aᵣ or Mᵣ incorrectly (e.g., dividing by Mᵣ for an element).
  • Rounding intermediate values too early and distorting the ratio; keep guard digits until the final answer.

B. Unit Mistakes

  • Forgetting to convert cm³ to dm³ for concentration: V(dm³) = V(cm³)/1000.
  • Using 24 dm³/mol at RTP but leaving the answer in cm³ without converting.

5. Exam Tips

Mark-Scheme Language

Use precise language: “used up completely”, “in excess”, and “therefore limits the amount of product”. Include the reason, not only the name of the limiting reactant.

  • Show the balanced equation before any calculations.
  • Write a clear limiting decision line (e.g., “moles ÷ coefficient is smaller for HCl, so HCl is limiting”).

6. Worked Examples

Modelled example 1

Mass + Moles Given (RTP Gas Volume)

Core

Problem

2.4 g of magnesium reacts with 0.15 mol of hydrochloric acid: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g). Identify the limiting reactant and find the volume of H₂ produced at RTP using 24 dm³ mol⁻¹.
Study the worked solution

  1. Put both reactants in moles

    Method

    Convert magnesium mass to moles and retain the supplied acid amount.

    Reason

    Reactants must be compared as amounts, not as unlike mass and mole quantities.

    Working

    n(Mg) = 2.4/24 = 0.10 mol; n(HCl) = 0.15 mol.

  2. Compare moles per coefficient

    Method

    Divide each amount by its balanced-equation coefficient.

    Reason

    The smaller reaction extent identifies the reactant used up first.

    Working

    Mg:0.10/1 = 0.10; HCl:0.15/2 = 0.075. Therefore HCl is limiting.

  3. Find hydrogen amount

    Method

    Use the 2:1 ratio from HCl to H₂.

    Reason

    Product must be calculated from the limiting reactant.

    Working

    n(H₂) = 0.15/2 = 0.075 mol.

  4. Convert amount to RTP volume

    Method

    Multiply hydrogen moles by 24 dm³ mol⁻¹.

    Reason

    This is the molar gas volume at RTP.

    Working

    V(H₂) = 0.075(24) = 1.8 dm³.

Guided practice 2

Solution + Solid (Gas Volume at RTP)

About 9 min

Problem

For Na₂CO₃(s) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g), 25.0 cm³ of 1.0 mol dm⁻³ HCl reacts with 2.65 g Na₂CO₃. Identify the limiting reactant and find the volume of CO₂ produced at RTP.

Convert, compare, then calculate product

Limiting reactant

Hints

Hint 1: convert both reactants
n(HCl) = cV using 0.0250 dm³; n(Na₂CO₃) = m/M using M = 106.
Hint 2: include coefficients
Compare 0.0250/2 for HCl with 0.0250/1 for sodium carbonate.
View solution step by step

  1. Calculate both reactant amounts

    Method

    Use cV for acid and m/M for the solid.

    Reason

    Both reactants must be expressed in moles before comparison.

    Working

    n(HCl) = 1.0(0.0250) = 0.0250 mol; n(Na₂CO₃) = 2.65/106 = 0.0250 mol.

  2. Apply coefficients

    Method

    Divide each amount by its equation coefficient.

    Reason

    Equal mole amounts are not stoichiometric here because the required ratio is 1:2.

    Working

    Na₂CO₃:0.0250/1 = 0.0250; HCl:0.0250/2 = 0.0125, so HCl is limiting.

  3. Find carbon dioxide amount

    Method

    Use the 2:1 acid-to-carbon-dioxide ratio.

    Reason

    Product is governed by the limiting acid amount.

    Working

    n(CO₂) = 0.0250/2 = 0.0125 mol.

  4. Calculate gas volume

    Working

    V(CO₂) = 0.0125(24) = 0.300 dm³ = 300 cm³.

Common misconception 3

Smaller Mass Is Not the Limiting Test

Find and correct the mistake

Learner response

For Mg + 2HCl → MgCl₂ + H₂, 2.4 g Mg reacts with 0.15 mol HCl. A student says magnesium must be limiting because 2.4 is the smaller given number. Explain the reasoning error and identify the limiting reactant.

Use comparable stoichiometric quantities

Valid comparison
Limiting reactant

View solution step by step

  1. Reject the raw-number comparison

    Method

    Do not compare 2.4 g directly with 0.15 mol.

    Reason

    The quantities have different units and neither accounts for the equation coefficients.

    Working

    First convert: n(Mg) = 2.4/24 = 0.10 mol; n(HCl) = 0.15 mol.

  2. Use the stoichiometric test

    Method

    Compare moles per coefficient.

    Reason

    The smaller value corresponds to the reactant used up first.

    Working

    0.10/1 = 0.10 for Mg; 0.15/2 = 0.075 for HCl. Therefore HCl is limiting.

Examiner practice 4

Two Masses Given (Find Excess Left Over)

8 marks

Examination question

For 2H₂(g) + O₂(g) → 2H₂O(l), 5.0 g of H₂ reacts with 32.0 g of O₂. Identify the limiting reactant, find the mass of water formed, and find the mass of excess reactant left. [8 marks]

Show the limiting, product and leftover chains

View solution step by step

  1. Calculate hydrogen moles

    1 mark

    Method

    Divide hydrogen mass by its molar mass.

    Reason

    Reactant amounts must be in moles before applying the equation ratio.

    Working

    n(H₂) = 5.0/2 = 2.5 mol.

  2. Calculate oxygen moles

    1 mark

    Method

    Divide oxygen mass by its molar mass.

    Reason

    This gives a quantity directly comparable with the hydrogen amount.

    Working

    n(O₂) = 32.0/32 = 1.0 mol.

  3. Identify the limiting reactant

    2 marks

    Method

    Compare moles per coefficient and state the decision.

    Reason

    The smaller reaction extent is used up first.

    Working

    H₂:2.5/2 = 1.25; O₂:1.0/1 = 1.0, so O₂ is limiting.

  4. Find water mass

    2 marks

    Method

    Use the 1:2 oxygen-to-water ratio, then multiply by water molar mass.

    Reason

    One mole of limiting O₂ produces two moles of water.

    Working

    n(H₂O) = 2.0 mol; m(H₂O) = 2.0(18) = 36 g.

  5. Find excess hydrogen left

    2 marks

    Method

    Subtract hydrogen used from hydrogen supplied, then convert to mass.

    Reason

    Two moles of hydrogen react with the one mole of limiting oxygen.

    Working

    n(H₂)_left = 2.5-2.0 = 0.5 mol; m_left = 0.5(2) = 1.0 g.

Challenge 5

Gas Volume + Solid (Mass of Product)

Minimal support

Representation transfer

For Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g), 16.0 g of Fe₂O₃ reacts with 11.2 dm³ of CO at RTP. Identify the limiting reactant and find the mass of iron produced. Use Aᵣ: Fe = 56, O = 16.

Convert solid mass and gas volume to moles

Limiting reactant

Hints

Hint 1: two conversion routes
Use n = m/M for Fe₂O₃ and n = V/24 for CO.
Hint 2: compare equation extents
Compare 0.100/1 with (11.2/24)/3.
View solution step by step

  1. Convert both reactants

    Method

    Convert solid mass and RTP gas volume into moles.

    Reason

    A common amount unit is required for stoichiometric comparison.

    Working

    n(Fe₂O₃) = 16.0/160 = 0.100 mol; n(CO) = 11.2/24 = 0.467 mol.

  2. Identify the limiting reactant

    Method

    Divide by the coefficients 1 and 3.

    Reason

    The smaller reaction extent controls the product amount.

    Working

    0.100/1 = 0.100 for Fe₂O₃; 0.467/3 = 0.156 for CO, so Fe₂O₃ is limiting.

  3. Find iron amount

    Method

    Use the 1:2 ratio from Fe₂O₃ to Fe.

    Reason

    One mole of limiting oxide produces two moles of iron.

    Working

    n(Fe) = 2(0.100) = 0.200 mol.

  4. Convert to product mass

    Working

    m(Fe) = 0.200(56) = 11.2 g.

7. Mind Stretchers

Mind stretcher 1: Find Limiting Reactant and Leftover ConcentrationExtension

50.0 cm³ of 0.500 mol/dm³ NaOH reacts with 25.0 cm³ of 0.800 mol/dm³ H₂SO₄: H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)

  1. Identify the limiting reactant.
  2. Calculate the moles of Na₂SO₄ formed.
  3. Find the concentration of excess reactant remaining after mixing (total volume = 75.0 cm³).
Show Answer

Moles:

n(NaOH) = 0.500 × 0.0500 = 0.0250 mol; n(H₂SO₄) = 0.800 × 0.0250 = 0.0200 mol

Compare moles ÷ coefficient:

  • NaOH: 0.0250/2 = 0.0125
  • H₂SO₄: 0.0200/1 = 0.0200

Since 0.0125 < 0.0200, NaOH is limiting.

Moles of Na₂SO₄ formed = moles of H₂SO₄ used = 0.0125 mol.

Excess H₂SO₄ left: 0.0200-0.0125 = 0.0075 mol.

Total volume = 0.0750 dm³, so concentration left: 0.0075/0.0750 = 0.100 mol/dm³.

Mind stretcher 2: Precipitation Mass (1:1 Ratio but Still Limiting)Extension

BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq) 50.0 cm³ of 0.200 mol/dm³ BaCl₂ is mixed with 25.0 cm³ of 0.300 mol/dm³ Na₂SO₄.

  1. Identify the limiting reactant.
  2. Calculate the mass of BaSO₄ formed. (Use Mᵣ(BaSO₄) = 233.)
Show Answer

Moles:

n(BaCl₂) = 0.200 × 0.0500 = 0.0100 mol; n(Na₂SO₄) = 0.300 × 0.0250 = 0.00750 mol

Ratio is 1:1, so the smaller moles is limiting: Na₂SO₄ is limiting.

Moles of BaSO₄ formed = 0.00750 mol.

Mass = 0.00750 × 233 = 1.75 g.

8. Quiz

Quiz Time!

Ready to test your knowledge? Try the quiz and see if you can spot limiting reactants quickly under exam timing.

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