Mini Chemistry
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Stoichiometric Reacting Quantities

Establish a balanced equation, convert a supplied quantity to moles, apply its coefficient ratio and convert to the requested quantity.

  • 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 establish a correct balanced equation, convert the supplied quantity to amount, use the coefficient ratio and convert the required amount to the requested quantity.

The flexible pathway is:

supplied quantity → supplied moles → (balanced coefficient ratio)required moles → requested quantity.

It is a reasoning pathway, not a mnemonic: some same-condition gas-volume questions can use the coefficient ratio directly because equal gas volumes contain proportional amounts under the same conditions.

2. Modelled example

Modelled example 1

Make equation correctness the first calculation step

Core

Problem

Calculate the mass of aluminium oxide formed when 5.40 g aluminium reacts completely with oxygen. Use Al = 27.0, O = 16.0.
View solution step by step

  1. Establish the relationship

    Method

    Write correct formulae and balance before using numbers.

    Reason

    Aluminium oxide is Al₂O₃; coefficients 4:3:2 conserve atoms and define the mole ratio.

    Working

    4Al(s) + 3O₂(g) → 2Al₂O₃(s)

  2. Convert the supplied mass

    Method

    Divide the aluminium mass by its molar mass.

    Reason

    Equation coefficients compare moles, not grams.

    Working

    n(Al) = 5.40/27.0 = 0.200 mol

  3. Apply the mole ratio

    Method

    Multiply aluminium amount by target coefficient over source coefficient.

    Reason

    4 mol Al form 2 mol Al₂O₃.

    Working

    n(Al₂O₃) = 0.200 × 2/4 = 0.100 mol

  4. Convert to the requested mass

    Method

    Calculate product molar mass and multiply it by product amount.

    Reason

    The requested quantity is mass, so m = nM completes the pathway.

    Working

    M(Al₂O₃) = 2(27.0) + 3(16.0) = 102.0 g mol⁻¹; therefore m = 0.100(102.0) = 10.2 g.

3. Guided practice

Guided practice 2

Name the reason for each conversion

About 6 min

Problem

Calcium carbonate decomposes to calcium oxide and carbon dioxide. Calculate the mass of CO₂ from 25.0 g CaCO₃. Use M(CaCO₃) = 100.0 g mol⁻¹ and M(CO₂) = 44.0 g mol⁻¹.

Equation, amount, ratio, requested mass

Hints

Hint 1: equation first
Write the decomposition equation and check whether it is already balanced.
Hint 2: quantity pathway
Convert carbonate mass to amount before using the 1:1 coefficient ratio.
View solution step by step

  1. Equation first

    Method

    Write correct formulae, states and coefficients before using the data.

    Reason

    The balanced equation establishes the mole relationship used later.

    Working

    CaCO₃(s) → CaO(s) + CO₂(g)

  2. Known amount

    Method

    Divide carbonate mass by carbonate molar mass.

    Reason

    The equation ratio acts on amounts rather than masses.

    Working

    n(CaCO₃) = 25.0/100.0 = 0.250 mol.

  3. Ratio and target

    Method

    Use the 1:1 ratio, then multiply product amount by product molar mass.

    Reason

    The first step reaches carbon-dioxide amount; the second reaches the requested mass.

    Working

    The 1:1 ratio gives 0.250 mol CO₂, so m = 0.250(44.0) = 11.0 g.

4. Plausible error contrast

For N₂ + 3H₂ → 2NH₃, a learner converts 6.00 mol H₂ to 18.0 mol NH₃ by multiplying by 3. The first invalid step is reading the ratio backwards. The relevant coefficients are H₂:NH₃ = 3:2, so

n(NH₃) = 6.00 × 2/3 = 4.00 mol.

5. Changed-context transfer

Mind stretcher 1: Decide when a direct volume ratio is validExtension

2CO(g) + O₂(g) → 2CO₂(g). At the same temperature and pressure, what volume of oxygen reacts with 150 cm³ carbon monoxide, and what volume of carbon dioxide forms?

Show feedback

All species are gases at the same conditions, so volume follows the 2:1:2 coefficient ratio. Oxygen volume is 150(1/2) = 75.0 cm³ and carbon-dioxide volume is 150(2/2) = 150 cm³. No molar-volume conversion is needed for this ratio comparison.

6. Independent evidence

Zinc reacts with hydrochloric acid to form zinc chloride and hydrogen. Without being given a method name, calculate the mass of zinc chloride formed from 6.54 g zinc when acid is in excess. Use Zn = 65.4 and Cl = 35.5. Your evidence must show the balanced equation with states, conversion to zinc moles, the zinc-to-zinc-chloride ratio, conversion to the requested mass, units and a final magnitude check.

Continue to Limiting and Excess Reactants to decide which supplied reactant actually controls a reaction.