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How do we calculate the mass of a product from the mass of a reactant?

Reacting masses; using moles to calculate the mass of a product or reactant; limiting reactants; and deducing balanced equations from masses.

A focused answer to AQA GCSE Chemistry 4.3.2 and 4.3.3, covering how to calculate reacting masses using moles and mole ratios, the idea of a limiting reactant, and using masses to deduce the balancing numbers in an equation.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Calculating reacting masses
Limiting reactants (Higher)
Deducing balancing numbers (Higher)
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What this dot point is asking

AQA wants you to calculate the mass of a product or reactant using moles and the mole ratio from a balanced equation, identify and use the limiting reactant, and deduce the balancing numbers of an equation from reacting masses. This is the workhorse calculation of the topic and appears on almost every Paper 1. The limiting reactant and equation-deduction work is Higher tier.

Calculating reacting masses

Follow the three-step method, which works for any equation:

Convert the known mass to moles: $\text{moles} = \dfrac{\text{mass}}{M_r}$ .
Use the mole ratio from the balanced equation.
Convert the result back to mass: $\text{mass} = \text{moles} \times M_r$ .

The same three steps work whether you measure in grams, kilograms or tonnes, because the mole ratio is a pure number. Industrial calculations (extracting iron, making ammonia) are usually set in tonnes but solved identically.

Limiting reactants (Higher)

Adding more of the limiting reactant makes more product; adding more of the reactant in excess makes no difference to the amount of product. To find the limiting reactant, work out the moles of each reactant and compare them with the mole ratio in the equation. The reactant that provides fewer moles relative to the ratio is limiting, which is not always the one with the smaller mass.

Deducing balancing numbers (Higher)

If you know the masses that react, convert each to moles and find the simplest whole-number ratio of moles. This ratio gives the balancing numbers in the equation. For example, if $4.6$ g of sodium ( $A_r = 23$ , so $0.2$ mol) reacts with $1.6$ g of oxygen ( $M_r = 32$ , so $0.05$ mol), the Na to $O_2$ ratio is $0.2 : 0.05 = 4 : 1$ , giving $4Na + O_2 \rightarrow 2Na_2O$ .

Mass of product, limiting reactant included

$12$ g of magnesium is added to $200$ cm $^3$ of hydrochloric acid containing $0.30$ mol of $HCl$ : $Mg + 2HCl \rightarrow MgCl_2 + H_2$ . Find the mass of hydrogen produced. ( $A_r$ : Mg = 24, H = 1.)

step 1 Find the moles of each reactant

Moles of Mg $= 12 / 24 = 0.50$ mol. Moles of $HCl = 0.30$ mol (given).

step 2 Identify the limiting reactant

The ratio of Mg to $HCl$ is $1:2$ , so $0.50$ mol of Mg would need $1.0$ mol of $HCl$ . Only $0.30$ mol of $HCl$ is present, so $HCl$ is limiting and magnesium is in excess.

step 3 Use the limiting reactant in the ratio

The ratio of $HCl$ to $H_2$ is $2:1$ , so moles of $H_2 = 0.30 / 2 = 0.15$ mol.

step 4 Convert to mass

$M_r$ of $H_2 = 2$ , so mass of hydrogen $= 0.15 \times 2 = 0.30$ g.

Try this

Q1. For $C + O_2 \rightarrow CO_2$ , calculate the mass of $CO_2$ made from $6$ g of carbon ( $A_r$ C = 12, $M_r$ CO2 = 44). [3 marks]

Cue. Moles C $= 6 / 12 = 0.5$ ; ratio $1:1$ ; mass CO2 $= 0.5 \times 44 = 22$ g.

Q2. Define the limiting reactant. [1 mark]

Cue. The reactant that is used up first and so controls the amount of product.

Q3. $5.6$ g of iron ( $A_r = 56$ ) reacts completely with sulfur to make iron sulfide, $FeS$ . Calculate the mass of sulfur that reacts ( $A_r$ S = 32). [3 marks]

Cue. Moles Fe $= 5.6 / 56 = 0.10$ ; ratio Fe to S is $1:1$ , so $0.10$ mol of S; mass $= 0.10 \times 32 = 3.2$ g.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20194 marksIron is extracted in the blast furnace by the reaction

Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2

. Calculate the maximum mass of iron that can be extracted from

160

tonnes of iron(III) oxide. Relative atomic masses: Fe = 56, O = 16.

Show worked answer →

A 4-mark Higher reacting-mass calculation worked in tonnes (the method is identical to grams).

The $M_r$ of $Fe_2O_3 = (2 \times 56) + (3 \times 16) = 160$ (1 mark). "Moles" $= 160 / 160 = 1.0$ (1 mark, treating tonnes consistently). The ratio of $Fe_2O_3$ to $Fe$ is $1:2$ , so amount of iron $= 2.0$ (1 mark). Mass of iron $= 2.0 \times 56 = 112$ tonnes (1 mark).

Markers accept working in tonnes throughout since the ratios are unitless; the key error is using a $1:1$ ratio instead of $1:2$ .

AQA 20225 marksHydrogen reacts with nitrogen to make ammonia:

N_2 + 3H_2 \rightarrow 2NH_3

. A reaction mixture contains

28

g of nitrogen and

9.0

g of hydrogen. Determine which reactant is limiting, and calculate the maximum mass of ammonia that can form. Relative atomic masses: N = 14, H = 1.

Show worked answer →

A 5-mark Higher limiting-reactant question.

Moles of $N_2 = 28 / 28 = 1.0$ mol (1 mark). Moles of $H_2 = 9.0 / 2 = 4.5$ mol (1 mark). The equation needs $3$ mol of $H_2$ per mol of $N_2$ , so $1.0$ mol of $N_2$ needs only $3.0$ mol of $H_2$ . There is $4.5$ mol of $H_2$ available, so hydrogen is in excess and nitrogen is limiting (1 mark). Moles of $NH_3 = 2 \times 1.0 = 2.0$ mol (1 mark). $M_r$ of $NH_3 = 17$ , so mass $= 2.0 \times 17 = 34$ g (1 mark).

Markers reward comparing moles against the ratio, not comparing raw masses.

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Sources & how we know this

AQA GCSE Chemistry (8462) specification — AQA (2016)