Reacting Masses

This lesson teaches you how to use a balanced chemical equation to calculate the masses of reactants and products involved in a reaction. These reacting mass calculations are among the most common quantitative questions in GCSE Chemistry exams and are required by the Edexcel specification (1CH0).

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The Step-by-Step Method

Every reacting mass calculation follows the same systematic method. Learn this method and you can tackle any question.

flowchart TD
    A["Step 1: Write the balanced equation"] --> B["Step 2: Calculate Mr of each relevant substance"]
    B --> C["Step 3: Find the moles of the substance you know"]
    C --> D["Step 4: Use the molar ratio from the equation"]
    D --> E["Step 5: Calculate the mass of the substance you want"]
    style A fill:#e6f3ff,stroke:#333
    style B fill:#e6f3ff,stroke:#333
    style C fill:#e6f3ff,stroke:#333
    style D fill:#e6f3ff,stroke:#333
    style E fill:#e6f3ff,stroke:#333

The Method in Detail

1. Write the balanced equation for the reaction.
2. Calculate Mr of the substance you know about and the substance you want to find.
3. Calculate moles of the substance you know: moles = mass ÷ Mr.
4. Use the molar ratio from the balanced equation to find moles of the substance you want.
5. Calculate the mass of the substance you want: mass = moles × Mr.

Exam Tip: Even if you cannot complete the whole calculation, writing out the balanced equation and showing Mr values will earn you marks. Always show every step.

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Worked Examples

Example 1: Magnesium and Oxygen

Question: What mass of magnesium oxide is produced when 4.8 g of magnesium burns in excess oxygen?

Step 1: Balanced equation:

2Mg + O₂ → 2MgO

Step 2: Calculate Mr values:

• Mr of Mg = 24
• Mr of MgO = 24 + 16 = 40

Step 3: Calculate moles of Mg:

• moles = mass ÷ Mr = 4.8 ÷ 24 = 0.2 mol

Step 4: Use the molar ratio:

• From the equation: 2Mg → 2MgO (ratio is 1:1)
• So 0.2 mol Mg produces 0.2 mol MgO

Step 5: Calculate mass of MgO:

• mass = moles × Mr = 0.2 × 40 = 8.0 g

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Example 2: Calcium Carbonate Decomposition

Question: What mass of calcium oxide is produced when 50 g of calcium carbonate is heated?

Step 1: Balanced equation:

CaCO₃ → CaO + CO₂

Step 2: Calculate Mr values:

• Mr of CaCO₃ = 40 + 12 + (3 × 16) = 100
• Mr of CaO = 40 + 16 = 56

Step 3: Calculate moles of CaCO₃:

• moles = 50 ÷ 100 = 0.5 mol

Step 4: Use the molar ratio:

• 1 CaCO₃ → 1 CaO (ratio is 1:1)
• So 0.5 mol CaCO₃ produces 0.5 mol CaO

Step 5: Calculate mass of CaO:

• mass = 0.5 × 56 = 28 g

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Example 3: Reaction with a Different Ratio

Question: What mass of sodium hydroxide is needed to neutralise 7.3 g of hydrochloric acid?

Step 1: Balanced equation:

NaOH + HCl → NaCl + H₂O

Step 2: Calculate Mr values:

• Mr of HCl = 1 + 35.5 = 36.5
• Mr of NaOH = 23 + 16 + 1 = 40

Step 3: Calculate moles of HCl:

• moles = 7.3 ÷ 36.5 = 0.2 mol

Step 4: Use the molar ratio:

• 1 NaOH : 1 HCl (ratio is 1:1)
• So 0.2 mol HCl requires 0.2 mol NaOH

Step 5: Calculate mass of NaOH:

• mass = 0.2 × 40 = 8.0 g

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Example 4: A 2:1 Ratio

Question: What mass of hydrogen is produced when 11.5 g of sodium reacts with excess water?

Step 1: Balanced equation:

2Na + 2H₂O → 2NaOH + H₂

Step 2: Calculate Mr values:

• Mr of Na = 23
• Mr of H₂ = 2

Step 3: Calculate moles of Na:

• moles = 11.5 ÷ 23 = 0.5 mol

Step 4: Use the molar ratio:

• 2Na : 1H₂ (ratio is 2:1)
• So 0.5 mol Na produces 0.5 ÷ 2 = 0.25 mol H₂

Step 5: Calculate mass of H₂:

• mass = 0.25 × 2 = 0.5 g

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Example 5: Finding the Mass of a Reactant

Question: What mass of iron is needed to produce 32 g of iron(III) oxide? (Ar: Fe = 56, O = 16)

Step 1: Balanced equation:

4Fe + 3O₂ → 2Fe₂O₃

Step 2: Calculate Mr values:

• Mr of Fe = 56
• Mr of Fe₂O₃ = (2 × 56) + (3 × 16) = 112 + 48 = 160

Step 3: Calculate moles of Fe₂O₃:

• moles = 32 ÷ 160 = 0.2 mol

Step 4: Use the molar ratio:

• 4Fe : 2Fe₂O₃ (ratio is 4:2 = 2:1)
• So 0.2 mol Fe₂O₃ requires 0.2 × 2 = 0.4 mol Fe

Step 5: Calculate mass of Fe:

• mass = 0.4 × 56 = 22.4 g

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Excess and Limiting Reagent

In many reactions, one reactant is used in excess — there is more than enough of it. The other reactant is the limiting reagent — it is completely used up and determines how much product is formed.

Key Points

• The limiting reagent controls the amount of product formed. Once it is used up, the reaction stops.
• The excess reagent is left over at the end of the reaction.
• In calculations, you calculate using the limiting reagent (the one you know the exact amount of).

How to Identify the Limiting Reagent

Question: 4.8 g of magnesium is reacted with 8.0 g of oxygen. Which is the limiting reagent?

2Mg + O₂ → 2MgO