Mini Chemistry
Part of Mini EducationSwitch subject

Reacting Masses and Limiting Reagent

Find the limiting reagent and calculate the product yield and the reactant left over.

  • GCE A-Level H2 Chemistry 9476-2027

Reading settings

On this page

Limiting reagent questions are ratio questions. Your marks come from showing the chain clearly: balanced equation → moles → compare with coefficients → limiting reagent → product/leftover.

Build on the mole and Avogadro constant, using your course’s Stoichiometry topic navigation to review them when needed. Keep unit conversions and mole ratios together in your working.

Definitions (Must Know)

A. Limiting reagent

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

B. Theoretical yield

The theoretical yield is the maximum amount of product predicted by the balanced equation (assuming the limiting reagent reacts completely).

C. Excess reagent

An excess reagent is a reactant that is present in more than the stoichiometric amount, so some is left unreacted after the reaction finishes.

Key Ideas (What Earns Marks)

  • Convert reactant amounts to moles first.
  • Compare using the balanced equation (mole ratio).
  • Product moles are determined by the limiting reactant only.
  • If asked for leftover: find moles used from the limiting reactant + ratio, then subtract from initial.
Quick Recall (Fast Limiting Check)
  • Method 1: compare n/coefficient for each reactant (smallest is limiting).
  • Method 2: calculate “moles of product possible” from each reactant (smallest wins).
Limiting Reagent: Compare n ÷ Coefficient (Example)Example (Mg + 2HCl): n(Mg) = 0.100 mol, n(HCl) = 0.150 mol, so n/coeff is 0.100 vs 0.075; the smaller value identifies HCl as limiting.Limiting Reagent: Compare n ÷ Coefficient (Example)Reactant (adjusted)n ÷ coefficient (mol)
Example (Mg + 2HCl): n(Mg) = 0.100 mol, n(HCl) = 0.150 mol, so n/coeff is 0.100 vs 0.075; the smaller value identifies HCl as limiting.
Data table
Reactant (adjusted)n/coeff
Mg (n/1)0.1
HCl (n/2)0.075

Detailed Explanations

A. Limiting reagent workflow

Because the balanced equation fixes the mole ratio, the limiting reagent is the reactant that cannot supply enough moles to satisfy that ratio.

  1. Write a balanced equation.
  2. Convert all reactant amounts to moles.
  3. Compare actual moles to required ratio (or compute product moles from each reactant and choose the smaller).
  4. Use limiting reactant to find product moles, then convert to mass / volume / concentration.

Mini example: Mg + 2HCl → MgCl₂ + H₂

  • If n(Mg) = 0.100 but n(HCl) = 0.150, then 0.100 mol Mg would need 0.200 mol HCl → HCl is limiting.

B. Leftover reactant

If asked, calculate:

  • moles used (from limiting reactant and ratio)
  • moles left = moles initial − moles used

C. Fast method: compare n ÷ coefficient

For a reaction aA + bB → …:

  • compute n(A)/a and n(B)/b
  • the smaller value is limiting

Change the mass of magnesium and the volume and concentration of acid, then run the reaction to test your prediction of the limiting reagent and the volume of hydrogen.

t = 0 s

0.192 g of magnesium and 10 cm³ of 0.6 mol/dm³ hydrochloric acid. Amounts now: Mg 0.00790 mol, HCl 0.00600 mol, MgCl₂ 0 mol, H₂ 0 mol.

n(Mg)
0.00790 mol
n(HCl)
0.00600 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)

Worked Examples

Modelled example 1

Identify the Limiting Reagent

Core

Problem

Magnesium reacts with hydrochloric acid: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g) If 2.40 g magnesium reacts with 0.150 mol hydrochloric acid, identify the limiting reagent. Use Aᵣ(Mg) = 24.0.

Study the worked solution

  1. Convert magnesium to amount

    Method

    Divide magnesium mass by its molar mass.

    Reason

    The equation compares reacting amounts rather than masses.

    Working

    n(Mg) = 2.40/24.0 = 0.100 mol

  2. Test the acid requirement

    Reason

    The equation requires two moles of HCl for every mole of Mg.

    Working

    n(HCl\ required) = 2(0.100) = 0.200 mol

  3. Compare with the supply

    Method

    Compare 0.150 mol available with 0.200 mol required.

    Reason

    The acid runs out before all magnesium can react.

    Working

    0.150 < 0.200, so HCl is the limiting reagent.

Guided practice 2

Calculate Product Mass from the Limiting Reagent

About 6 min

Problem

Using the same reaction and amounts, calculate the mass of MgCl₂ formed. Use Aᵣ(Mg) = 24.0 and Aᵣ(Cl) = 35.5.

Try this before viewing the solution

Unit: g

Hints

Hint 1: start from the limiting reagent

Use the 0.150 mol HCl supply rather than the initial magnesium amount.

Hint 2: apply the product ratio

Two moles of HCl form one mole of MgCl₂.

View solution step by step

  1. Find product amount

    Method

    Divide the limiting HCl amount by two.

    Reason

    The balanced equation gives a 2:1 HCl-to-MgCl₂ ratio.

    Working

    n(MgCl₂) = 0.150/2 = 0.0750 mol

  2. Find molar mass

    Reason

    The product formula contains one Mg and two Cl atoms.

    Working

    M(MgCl₂) = 24.0 + 2(35.5) = 95.0 g mol⁻¹

  3. Convert to mass

    Method

    Use m = nM.

    Reason

    The calculated product amount is the maximum allowed by the limiting reagent.

    Working

    m = (0.0750)(95.0) = 7.13 g

Common misconception 3

Correct a Direct-mass Comparison

Find and correct the mistake

Learner claim

For 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s), 5.60 g iron is mixed with 4.80 g oxygen. A learner says oxygen is limiting because 4.80 g < 5.60 g. Identify the first error and determine the limiting reagent. Use Aᵣ(Fe) = 56.0 and M(O₂) = 32.0 g mol⁻¹.

Diagnose before deciding

Limiting reagent

View solution step by step

  1. Convert both masses to amounts

    Method

    Calculate moles before comparing reactants.

    Reason

    Equal masses do not represent equal particle amounts, and the equation uses a 4:3 mole ratio.

    Working

    n(Fe) = 5.60/56.0 = 0.100 mol; n(O₂) = 4.80/32.0 = 0.150 mol.

  2. Normalise by coefficients

    Method

    Compare 0.100/4 with 0.150/3.

    Reason

    The smaller reaction extent reaches zero first.

    Working

    0.0250 < 0.0500, so iron is limiting.

Examiner practice 4

Calculate Unreacted Magnesium

4 marks

Problem

In the original magnesium–hydrochloric acid mixture, calculate the mass of magnesium left unreacted. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Use the limiting acid

    1 mark

    Method

    Start with 0.150 mol HCl.

    Reason

    The limiting reagent determines how much magnesium can react.

    Working

    n(HCl) = 0.150 mol

  2. Find magnesium used

    1 mark

    Reason

    Two moles of HCl react with one mole of Mg.

    Working

    n(Mg\ used) = 0.150/2 = 0.0750 mol

  3. Find magnesium left

    1 mark

    Reason

    Excess remaining equals initial amount minus amount consumed.

    Working

    n(Mg\ left) = 0.100-0.0750 = 0.0250 mol

  4. Convert to mass

    1 mark

    Method

    Multiply the remaining amount by 24.0 g mol⁻¹.

    Reason

    The question asks for mass rather than amount.

    Working

    m = (0.0250)(24.0) = 0.600 g

Challenge 5

Analyse an Ammonia Batch

Minimal support

Problem

In a batch calculation, 14.0 kg nitrogen is mixed with 2.00 kg hydrogen: N₂(g) + 3H₂(g) → 2NH₃(g) Assuming complete reaction, identify the limiting reagent and calculate the theoretical ammonia mass and the mass of excess reactant left. Use M(N₂) = 28.0, M(H₂) = 2.00 and M(NH₃) = 17.0 g mol⁻¹.

Try this before viewing the solution

Hints

Hint 1: align mass units

Convert both kilogram masses to grams before using molar masses.

Hint 2: compare reaction extents

Compare n(N₂)/1 with n(H₂)/3.

View solution step by step

  1. Convert both reactants to amounts

    Method

    Convert kilograms to grams, then divide by molar mass.

    Reason

    Both reactants must be in the same amount basis before applying coefficients.

    Working

    n(N₂) = 14000/28.0 = 500 mol; n(H₂) = 2000/2.00 = 1000 mol.

  2. Identify the limiting reagent

    Reason

    500/1 = 500 reaction units are available from nitrogen, but 1000/3 = 333 from hydrogen.

    Working

    Hydrogen is limiting.

  3. Calculate ammonia mass

    Reason

    Three moles of hydrogen form two moles of ammonia.

    Working

    n(NH₃) = (2/3)(1000) = 667 mol; m = (667)(17.0) = 1.13 × 10⁴ g = 11.3 kg.

  4. Calculate nitrogen left

    Method

    Subtract nitrogen consumed from nitrogen supplied.

    Reason

    One mole of nitrogen reacts per three moles of limiting hydrogen.

    Working

    n(N₂\ used) = 1000/3 = 333 mol; n(N₂\ left) = 167 mol; m = (167)(28.0) = 4.67 kg.

Common Mistakes

  • Comparing masses directly instead of converting to moles.
  • Using the non-limiting reagent to calculate yield.
  • Forgetting to apply the stoichiometric ratio when finding moles used.

Use the topic check in Practise and check below to practise and check your understanding.

Exam Tips

  • A fast check: “How many moles of product does each reactant allow?” Choose the smaller.
  • Always write the mole ratio line explicitly (often a mark).

Mind Stretchers

Mind stretcher 1Extension

In the reaction 2CO + O₂ → 2CO₂, 0.80 mol of CO reacts with 0.30 mol of O₂. Find the moles of CO₂ formed.

Show Hint

Convert each reactant to moles and compare the available amount divided by its equation coefficient.

Show Answer

Mark scheme:

  • Need 1 mol O2 per 2 mol CO. For 0.80 mol CO, required O2 = 0.80/2 = 0.40 mol.
  • Available O2 = 0.30 mol, so O2 is limiting.
  • 1 mol O2 gives 2 mol CO2, so CO2 formed = 2(0.30) = 0.60 mol.

Mind stretcher 2: Using gas mass to determine purityExtension

Question. A 10.0 g impure sample of CaCO₃ produces 1.76 g of CO₂ with excess acid. Calculate the percentage purity. Use M(CO₂) = 44.0 and M(CaCO₃) = 100 g mol⁻¹.

Show Hint

CaCO₃ and CO₂ are in a 1:1 mole ratio.

Show Answer

n(CO₂) = 1.76/44.0 = 0.0400 mol, so the sample contained 0.0400 mol or 4.00 g of CaCO₃. Purity = (4.00/10.0) × 100 = 40.0%.

Syllabus and review details

Last reviewed: