Reacting Masses

This lesson covers how to calculate reacting masses using balanced equations and moles, as required by the AQA GCSE Combined Science Trilogy specification (8464). This is one of the most important calculation skills in GCSE chemistry.

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The Method for Reacting Mass Calculations

flowchart TD
    A["1. Write the balanced equation"] --> B["2. Calculate the Mr of the\nsubstances you need"]
    B --> C["3. Find the moles of the\nsubstance you know"]
    C --> D["4. Use the mole ratio from\nthe balanced equation"]
    D --> E["5. Calculate the mass of the\nsubstance you want"]
    style A fill:#3b82f6,color:#fff,stroke:#2563eb
    style C fill:#f59e0b,color:#000,stroke:#d97706
    style E fill:#22c55e,color:#fff,stroke:#16a34a

Step-by-Step

1. Write the balanced equation.
2. Calculate the $M_r$Mr​ of the relevant substances.
3. Find the moles of the substance whose mass you know: $\text{moles} = \dfrac{\text{mass}}{M_r}$moles=Mr​mass​
4. Use the mole ratio from the balanced equation to find the moles of the substance you want.
5. Calculate the mass of the substance you want: $\text{mass} = \text{moles} \times M_r$mass=moles×Mr​

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Worked Examples

Example 1 — Magnesium and Oxygen

Question: What mass of magnesium oxide (MgO) is produced when 4.8 g of magnesium reacts completely with oxygen?

$2\text{Mg}(s) + \text{O}_2(g) \rightarrow 2\text{MgO}(s)$2Mg(s)+O2​(g)→2MgO(s)

Step 1: $M_r$Mr​ of Mg = 24, $M_r$Mr​ of MgO = 24 + 16 = 40

Step 2: Moles of Mg: $\text{moles} = \frac{4.8}{24} = 0.2 \text{ mol}$moles=244.8​=0.2 mol

Step 3: Mole ratio — from the equation, 2 mol Mg produces 2 mol MgO (ratio 1:1). $\text{moles of MgO} = 0.2 \text{ mol}$moles of MgO=0.2 mol

Step 4: Mass of MgO: $\text{mass} = 0.2 \times 40 = 8.0 \text{ g}$mass=0.2×40=8.0 g

Example 2 — Thermal Decomposition of Calcium Carbonate

Question: What mass of calcium oxide is produced when 25 g of calcium carbonate is heated?

$\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)$CaCO3​(s)→CaO(s)+CO2​(g)

$M_r$Mr​ of $\text{CaCO}_3$CaCO3​ = 100, $M_r$Mr​ of CaO = 56

$\text{moles of CaCO}_3 = \frac{25}{100} = 0.25 \text{ mol}$moles of CaCO3​=10025​=0.25 mol

Ratio is 1:1, so moles of CaO = 0.25 mol.

$\text{mass of CaO} = 0.25 \times 56 = 14 \text{ g}$mass of CaO=0.25×56=14 g

Example 3 — Sodium Carbonate and Hydrochloric Acid

Question: What mass of CO₂ is produced when 10.6 g of $\text{Na}_2\text{CO}_3$Na2​CO3​ reacts with excess HCl?

$\text{Na}_2\text{CO}_3(s) + 2\text{HCl}(aq) \rightarrow 2\text{NaCl}(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)$Na2​CO3​(s)+2HCl(aq)→2NaCl(aq)+H2​O(l)+CO2​(g)

$M_r$Mr​ of $\text{Na}_2\text{CO}_3$Na2​CO3​ = 106, $M_r$Mr​ of $\text{CO}_2$CO2​ = 44

$\text{moles of Na}_2\text{CO}_3 = \frac{10.6}{106} = 0.1 \text{ mol}$moles of Na2​CO3​=10610.6​=0.1 mol

Ratio $\text{Na}_2\text{CO}_3 : \text{CO}_2 = 1 : 1$Na2​CO3​:CO2​=1:1

$\text{mass of CO}_2 = 0.1 \times 44 = 4.4 \text{ g}$mass of CO2​=0.1×44=4.4 g

Example 4 — Zinc and Hydrochloric Acid

Question: What mass of hydrogen gas is produced when 13 g of zinc reacts with excess hydrochloric acid?

$\text{Zn}(s) + 2\text{HCl}(aq) \rightarrow \text{ZnCl}_2(aq) + \text{H}_2(g)$Zn(s)+2HCl(aq)→ZnCl2​(aq)+H2​(g)

$A_r$Ar​ of Zn = 65, $M_r$Mr​ of $\text{H}_2$H2​ = 2

$\text{moles of Zn} = \frac{13}{65} = 0.2 \text{ mol}$moles of Zn=6513​=0.2 mol

Ratio Zn : $\text{H}_2$H2​ = 1 : 1

$\text{mass of H}_2 = 0.2 \times 2 = 0.4 \text{ g}$mass of H2​=0.2×2=0.4 g

Example 5 — Iron and Chlorine

Question: What mass of iron is needed to react completely with 21.3 g of $\text{Cl}_2$Cl2​?

$2\text{Fe}(s) + 3\text{Cl}_2(g) \rightarrow 2\text{FeCl}_3(s)$2Fe(s)+3Cl2​(g)→2FeCl3​(s)

$M_r$Mr​ of $\text{Cl}_2$Cl2​ = 71, $A_r$Ar​ of Fe = 56

$\text{moles of Cl}_2 = \frac{21.3}{71} = 0.3 \text{ mol}$moles of Cl2​=7121.3​=0.3 mol

Ratio Fe : $\text{Cl}_2$Cl2​ = 2 : 3, so: $\text{moles of Fe} = \frac{2}{3} \times 0.3 = 0.2 \text{ mol}$moles of Fe=32​×0.3=0.2 mol

$\text{mass of Fe} = 0.2 \times 56 = 11.2 \text{ g}$mass of Fe=0.2×56=11.2 g

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Understanding the Mole Ratio

The coefficients in a balanced equation tell you the mole ratio between reactants and products.

Equation	Ratio
$\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$CaCO3​→CaO+CO2​	1 : 1 : 1
$2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$2Mg+O2​→2MgO	2 : 1 : 2
$2\text{Fe} + 3\text{Cl}_2 \rightarrow 2\text{FeCl}_3$2Fe+3Cl2​→2FeCl3​	2 : 3 : 2
$\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$N2​+3H2​→2NH3​	1 : 3 : 2

Exam Tip (AQA 8464): Always write out the balanced equation first and clearly state the mole ratio. Even if you make an arithmetic error later, you can still pick up method marks.

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Common Mistakes to Avoid

Mistake	Correction
Forgetting to write the balanced equation	The equation gives you the mole ratio — without it you cannot calculate
Using masses instead of moles in the ratio	The coefficients give the mole ratio, not the mass ratio
Ignoring the mole ratio	Always check whether the ratio is 1:1, 1:2, 2:3, etc.
Not converting back to mass at the end	The final answer should be in grams unless stated otherwise

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Summary

• Reacting mass calculations use the equation: $\text{moles} = \dfrac{\text{mass}}{M_r}$moles=Mr​mass​
• The balanced equation gives the mole ratio between substances.
• Steps: balanced equation → $M_r$Mr​ → moles of known → mole ratio → moles of unknown → mass of unknown.
• Always show your working clearly and state the mole ratio.
• If a question says "excess", the other reactant is the one you use in your calculation.

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Extended Worked Examples — Reacting Masses Mastery

Reacting mass calculations appear on every AQA Trilogy Chemistry Paper 1. The structure is always the same: equation → $M_r$Mr​ → moles → mole ratio → moles of target → mass of target. Below are six further worked examples covering different reaction types.

Worked Example 6 — Decomposition of Sodium Hydrogencarbonate

Question: A chef uses 16.8 g of baking soda ($\text{NaHCO}_3$NaHCO3​). What mass of sodium carbonate remains after full decomposition?

$2\text{NaHCO}_3(s) \rightarrow \text{Na}_2\text{CO}_3(s) + \text{H}_2\text{O}(g) + \text{CO}_2(g)$2NaHCO3​(s)→Na2​CO3​(s)+H2​O(g)+CO2​(g)

$M_r$Mr​ of $\text{NaHCO}_3 = 23 + 1 + 12 + 48 = 84$NaHCO3​=23+1+12+48=84; $M_r$Mr​ of $\text{Na}_2\text{CO}_3 = 106$Na2​CO3​=106.

$n_{\text{NaHCO}_3} = \frac{16.8}{84} = 0.2 \text{ mol}$nNaHCO3​​=8416.8​=0.2 mol

Ratio 2 : 1 for $\text{NaHCO}_3 : \text{Na}_2\text{CO}_3$NaHCO3​:Na2​CO3​.

$n_{\text{Na}_2\text{CO}_3} = 0.1 \text{ mol}$nNa2​CO3​​=0.1 mol

$m_{\text{Na}_2\text{CO}_3} = 0.1 \times 106 = 10.6 \text{ g}$mNa2​CO3​​=0.1×106=10.6 g

Worked Example 7 — Acid and Metal Carbonate

Question: What mass of $\text{CO}_2$CO2​ is produced when 5.0 g of $\text{CaCO}_3$CaCO3​ reacts fully with excess HCl?

$\text{CaCO}_3(s) + 2\text{HCl}(aq) \rightarrow \text{CaCl}_2(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)$CaCO3​(s)+2HCl(aq)→CaCl2​(aq)+H2​O(l)+CO2​(g)

$M_r$Mr​ of $\text{CaCO}_3 = 100$CaCO3​=100.

$n_{\text{CaCO}_3} = \frac{5.0}{100} = 0.05 \text{ mol}$nCaCO3​​=1005.0​=0.05 mol

Ratio 1 : 1 for $\text{CaCO}_3 : \text{CO}_2$CaCO3​:CO2​.

$m_{\text{CO}_2} = 0.05 \times 44 = 2.2 \text{ g}$mCO2​​=0.05×44=2.2 g

Worked Example 8 — Iron Extraction from Blast Furnace

Question: What mass of iron is produced from 480 kg of $\text{Fe}_2\text{O}_3$Fe2​O3​?

$\text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \rightarrow 2\text{Fe}(l) + 3\text{CO}_2(g)$Fe2​O3​(s)+3CO(g)→2Fe(l)+3CO2​(g)

$M_r$Mr​ of $\text{Fe}_2\text{O}_3 = (2 \times 56) + (3 \times 16) = 160$Fe2​O3​=(2×56)+(3×16)=160.