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Limiting reactants and excess left over

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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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.

Why a reaction stops

A balanced equation tells you how many particles must react together. For example:

2H₂(g) + O₂(g) → 2H₂O(l)

Two hydrogen molecules need one oxygen molecule. If all the hydrogen is used up, the remaining oxygen cannot make more water on its own. The limiting reactant is completely used up and sets the maximum product amount. An excess reactant remains afterwards. We assume the stated reaction goes to completion; the water shown is after cooling.

Compare supplies with the two-to-one reacting ratioBefore reaction: four hydrogen molecules and three oxygen molecules. Two hydrogen molecules react with each oxygen molecule, so the supply of hydrogen supports two reacting groups while oxygen supports three. After complete reaction: four water molecules and one oxygen molecule left over, with no hydrogen molecules remaining.HHOOOHHBefore reaction4 H₂ molecules3 O₂ moleculesH₂: 4 ÷ 2 = 2 groupsO₂: 3 ÷ 1 = 3 groupsH₂ is limitingAfter complete reaction4 H₂O molecules formed1 O₂ molecule left over2 reacting groups × 2 H₂O= 4 H₂O moleculesO₂ is in excess
A counting model: four H₂ molecules react with two of the three O₂ molecules. Four H₂O molecules form and one O₂ molecule remains. The eight H atoms and six O atoms are conserved. Water is shown after cooling; molecular sizes, spacing and arrangement are schematic.

The diagram starts with more hydrogen molecules than oxygen molecules, yet hydrogen is limiting. Comparing particle counts alone misses the required 2:1 ratio.

The theoretical yield is the maximum product amount calculated from the balanced equation and the limiting reactant, assuming complete reaction with no competing reactions. The amount you actually collect may be smaller.

A reliable calculation method

  1. Write the balanced equation.
  2. Convert the supplied amounts to moles: use n = m/M for a mass, n = cV for a solution, or n = V/Vₘ for a gas.
  3. Divide each reactant’s mole amount by its coefficient. The smallest result identifies the limiting reactant.
  4. Use the limiting amount and the equation ratio to calculate the product.

Why divide by the coefficient? The result tells you how many complete reacting groups each supply can support. For the diagram, hydrogen supports 4/2 = 2 groups and oxygen supports 3/1 = 3. Only two groups can react: they form four water molecules and leave one oxygen molecule.

If both results are equal, the reactants are supplied in exactly the required ratio: both are used up together and neither is in excess.

Choose the correct amount conversion

Use the molar mass of the substance in the formula: M(Mg) = 24 g mol⁻¹, but M(O₂) = 32 g mol⁻¹, because oxygen gas contains molecules with two oxygen atoms. For n = cV with c in mol dm⁻³, convert cm³ to dm³ by dividing by 1000. For gases at RTP, use the syllabus approximation Vₘ = 24 dm³ mol⁻¹.

Revise mass to moles, gas volumes, solution concentration or balancing equations if needed.

When all the compared reactants are gases measured at the same temperature and pressure, their volume ratios follow their mole ratios. You may compare volume divided by coefficient directly in that case. Moles provide a general method when the supplied quantities include masses or solutions.

Another way to check your decision

Calculate how much product each reactant could make if the other reactants were in excess. The smaller possible product amount is the theoretical yield; the reactant that gives it is limiting. This is the same reasoning as the coefficient method.

Finding the excess left over

Use the limiting reactant to calculate how much of the excess reactant is used. Then subtract:

n_left = nᵢₙᵢₜᵢₐₗ-n_used

Convert the remaining mole amount to mass or concentration only if the question asks for it. A negative amount left over means the limiting decision or ratio needs checking.

Weigh out magnesium, choose the acid and watch the particles react. Compare each reactant’s n ÷ coefficient before you start, then check which one runs out.

t = 0 s

0.096 g of magnesium and 20 cm³ of 0.6 mol/dm³ hydrochloric acid. Amounts now: Mg 0.00395 mol, HCl 0.0120 mol, MgCl₂ 0 mol, H₂ 0 mol.

n(Mg)
0.00395 mol
n(HCl)
0.0120 mol
Volume of H2
0 cm³
Limiting reactant
—
Reaction
g
cm³
mol/dm³

Try this

0 of 4 done

  1. Choose magnesium and acid that react with nothing left over, then start. (not done yet)

  2. Run Mg + HCl once with the magnesium used up and once with the acid used up. (not done yet)

  3. Mix hydrogen and oxygen so that no gas is left after the spark. (not done yet)

  4. Weigh the crucible part-way through heating, then again once its mass stops changing. (not done yet)

Make the deciding comparison visible

Show both mole amounts and both amounts divided by coefficient, then state which reactant is limiting and why. A smaller mass or fewer moles alone does not establish the answer. Keep unrounded intermediate values, include units, and round the final answer to the precision supported by the data.

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. Use M(Mg) = 24 g mol⁻¹ and assume complete reaction with no gas lost.
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 amount divided by coefficient 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 1.00 mol dm⁻³ HCl reacts with 2.65 g Na₂CO₃. Identify the limiting reactant and find the volume of CO₂ produced at RTP. Use M(Na₂CO₃) = 106 g mol⁻¹ and Vₘ = 24 dm³ mol⁻¹. Assume pure reactants and complete reaction.

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 g mol⁻¹.

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) = 1.00(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

Fewer Moles Can Still Be in Excess

Find and correct the mistake

Learner response

For Mg + 2HCl → MgCl₂ + H₂, 0.050 mol Mg reacts with 0.060 mol HCl. A student says magnesium must be limiting because there are fewer moles of magnesium. 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

    The reactants are already expressed in moles; include the required ratio.

    Reason

    One mole of magnesium needs two moles of HCl, so comparing mole amounts alone ignores how much acid the magnesium needs.

    Working

    0.050 mol Mg would need 0.10 mol HCl, but only 0.060 mol HCl is available.

  2. Use the stoichiometric test

    Method

    Compare moles per coefficient.

    Reason

    The smaller value corresponds to the reactant used up first.

    Working

    0.050/1 = 0.050 for Mg; 0.060/2 = 0.030 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 pure H₂ reacts completely with 32.0 g of pure O₂ after ignition; the products are then cooled. Use molar masses in g mol⁻¹: H₂ = 2, O₂ = 32, H₂O = 18. 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 = 1.00 mol.

  3. Identify the limiting reactant

    2 marks

    Method

    Compare moles per coefficient and state the decision.

    Reason

    The reactant with the smaller amount divided by coefficient is used up first.

    Working

    H₂:2.5/2 = 1.25; O₂:1.00/1 = 1.00, 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) = 2.00 mol; m(H₂O) = 2.00(18) = 36.0 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 = 2.5-2.00 = 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 is heated with an amount of CO that occupies 11.2 dm³ at RTP before heating. Assume complete reduction by the stated reaction. 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; (11.2/24)/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.

Try independently

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 acid remaining. Assume complete neutralisation by the stated equation and a final solution volume of 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 = 0.00750 mol.

Total volume = 0.0750 dm³, so concentration left: 0.00750/0.0750 = 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 M(BaSO₄) = 233 g mol⁻¹ and assume complete precipitation.
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.

Unrounded mass = 0.00750 × 233 = 1.7475 g. To three significant figures, the mass is 1.75 g.

Mind stretcher 3: Exactly the required ratioExtension

For 2H₂ + O₂ → 2H₂O, 0.20 mol hydrogen and 0.10 mol oxygen react completely. How many moles of water form, and which reactant is left over?

Show answer

Both amounts divided by coefficient are 0.10 mol: 0.20/2 = 0.10/1. The reactants are in the exact 2:1 ratio, so 0.20 mol water forms and neither reactant remains. There is no excess reactant.

Practise and check

Practise choosing the limiting reactant, then calculating product and excess amounts. Next, percentage yield and purity connects these maximum amounts to experimental results.

Open the chemical calculations topic check.

Syllabus and review details

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