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Limiting and Excess Reactants

Compare available amounts with stoichiometric requirements, identify the limiting reactant and calculate maximum product and excess remaining.

  • SEC G3 Combined Science Chemistry component 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)

1. Outcome and prerequisites

By the end, you should be able to compare each available amount with the balanced stoichiometric requirement, identify limiting and excess reactants from evidence, calculate maximum product, and calculate excess remaining at G3 Science depth.

For each reactant, compare

(available amount)/coefficient.

The smaller value is the possible reaction extent and identifies the reactant that runs out first. Raw mass or raw mole values alone are not enough when coefficients differ.

2. Modelled example

Modelled example 1

Use both reactants to determine maximum product

Core

Problem

4.86 g Mg reacts with 10.95 g HCl. Identify the limiting reactant, calculate the maximum amount of hydrogen, and find the amount of excess reactant remaining. Use Mg = 24.3, H = 1.0, Cl = 35.5.
View solution step by step

  1. Establish the equation

    Method

    Write and balance the reaction before comparing supplies.

    Reason

    The requirement must come from the balanced chemical relationship.

    Working

    Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

  2. Convert both available masses

    Method

    Divide each reactant mass by its own molar mass.

    Reason

    Limiting-reactant evidence compares available amounts against coefficients, not raw masses.

    Working

    n(Mg) = 4.86/24.3 = 0.200 mol; M(HCl) = 36.5 g mol⁻¹, so n(HCl) = 10.95/36.5 = 0.300 mol.

  3. Compare amount per coefficient

    Method

    Normalise each amount to one reaction unit.

    Reason

    The equation needs twice as much HCl as Mg.

    Working

    0.200/1 = 0.200, 0.300/2 = 0.150 HCl is limiting.

  4. Calculate product and excess

    Method

    Use the limiting reaction extent for product and subtract the amount of excess reactant consumed.

    Reason

    The limiting amount controls maximum product and therefore how much excess reactant can react.

    Working

    n(H₂) = 0.150 mol. It consumes 0.150 mol Mg, leaving 0.200-0.150 = 0.0500 mol Mg.

3. Guided practice

Guided practice 2

Compare requirements rather than wording cues

About 6 min

Problem

0.80 mol N₂ is mixed with 1.80 mol H₂ for N₂ + 3H₂ → 2NH₃. Identify the limiting reactant and maximum ammonia amount.

Show both comparisons

Hints

Hint 1: normalise
Divide each available amount by its own equation coefficient.
Hint 2: product amount
Use the smaller normalised value as reaction extent, then multiply by the ammonia coefficient.
View solution step by step

  1. Normalise both amounts

    Method

    Divide each available amount by the reactant’s coefficient.

    Reason

    This compares how many complete reaction units each supply can support.

    Working

    0.80/1 = 0.80 for N₂; 1.80/3 = 0.600 for H₂.

  2. Use the smaller extent

    Method

    Select hydrogen as limiting and multiply reaction extent by the ammonia coefficient.

    Reason

    The smaller extent runs out first, and each reaction unit forms two moles of ammonia.

    Working

    H₂ is limiting and n(NH₃) = 0.600(2) = 1.20 mol.

4. Plausible error contrast

A learner correctly finds 0.100 mol Mg and 0.0400 mol HCl, then uses the initial magnesium amount to report 0.100 mol hydrogen. The first invalid step is using an amount from the excess reactant after HCl has been identified as limiting. With Mg + 2HCl → MgCl₂ + H₂, the maximum is 0.0400/2 = 0.0200 mol hydrogen.

Repair the controlling step

Do not restart every conversion when they are correct. Replace the first downstream use of an excess-reactant amount with the limiting-reactant amount and coefficient ratio.

5. Changed-context transfer

Mind stretcher 1: Find product mass and reactant remainingExtension

2Cu(s) + O₂(g) → 2CuO(s). A sealed vessel contains 0.500 mol Cu and 0.180 mol O₂. Determine the limiting reactant, maximum amount of CuO and amount of the excess reactant remaining.

Show feedback

0.500/2 = 0.250 while 0.180/1 = 0.180, so oxygen is limiting. It forms 2(0.180) = 0.360 mol CuO and consumes 0.360 mol Cu. Copper remaining is 0.500-0.360 = 0.140 mol.

6. Independent evidence

6.40 g sulfur dioxide reacts with 3.20 g oxygen to form sulfur trioxide. Use S = 32.0 and O = 16.0. Establish and balance the equation, convert both masses to amounts, compare each with its coefficient, identify limiting and excess reactants, calculate the maximum mass of sulfur trioxide and the mass of excess reactant remaining. Check that no product amount requires more of the limiting reactant than is available.

Continue to Gas-Volume Calculations.