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IGCSE · Cambridge (CIE) · 0654

Blast-furnace equations and oxygen accounting

Balance the five required equations and distinguish redox, decomposition and slag formation.

Section C9.6 · Extraction of metals

Extraction of metals · Cambridge IGCSE 0654 · C9.6

Mapped to the named 0654 section using teaching from the corresponding single-science course. A 0654-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade.

Before you begin

  • Explain all five furnace stages in words.
  • Balance equations and apply Extended reacting-mass ratios.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

How many iron atoms occur in Fe₂O₃?

What you will learn

  • State and balance the five required extraction equations.
  • Use them to check oxygen transfer and a supplied theoretical iron calculation.

Match each equation with its purpose

Coke combustion is C+O₂→CO₂. Carbon-monoxide formation is C+CO₂→2CO. Hematite reduction is Fe₂O₃+3CO→2Fe+3CO₂. The third equation shows iron oxide losing oxygen while carbon monoxide gains oxygen; matching oxygen atoms does not require molecular oxygen as a product.

Limestone decomposition is CaCO₃→CaO+CO₂. Slag formation is CaO+SiO₂→CaSiO₃. In the last reaction the basic calcium oxide reacts with acidic silicon dioxide; it is not the equation that produces metallic iron. Balance by coefficients while preserving the correct formulas and distinguish the chemical role of each equation.

Use the reduction ratio with explicit assumptions

One mole of Fe₂O₃ forms two moles of Fe in the balanced reduction. With supplied relative atomic masses Fe=56 and O=16, Fe₂O₃ has formula mass 160 and its iron contribution is 112. Thus a theoretical pure-hematite calculation uses an iron mass factor 112/160=0.70, assuming complete reduction and adequate carbon monoxide.

Ore impurities, incomplete conversion and product losses reduce actual recovery relative to the stated pure theoretical case. Do not use the same mass factor directly on an impure ore without first finding its hematite amount. The equation's three CO molecules per Fe₂O₃ formula unit explain oxygen acceptance; the three is not the number of iron atoms produced.

Sources for this lesson

Pause and explain

Which required equation forms slag?

Put the idea to work

Worked example

Using Fe=56 and O=16, find the theoretical iron from 160 g pure Fe₂O₃ with sufficient CO and complete reduction.

Show the worked solution
  1. Fe₂O₃ has molar mass 2×56+3×16=160 g/mol.
  2. 160 g is 1 mol and gives 2 mol Fe.
  3. Iron mass is 2×56=112 g under the stated complete-reduction assumptions.

Answer 112 g theoretical iron.

Try a topic test

Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.

Test this topic →

From guided practice to a new situation

Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.

1 · Guided

Write coke combustion and carbon-monoxide formation equations and state their roles.

Give me a hint

The carbon dioxide formed first is used with further hot carbon.

Compare my reasoning
  1. C+O₂→CO₂ describes combustion and heat release.
  2. C+CO₂→2CO describes carbon-monoxide formation.
  3. CO then serves as the reducing gas for hematite in the separate ore reaction.

C+O₂→CO₂; C+CO₂→2CO; heat first, reducing-gas formation next.

Look for these in your work

  • Both equations balance.
  • Their separate purposes are stated.
2 · Independent

Write the limestone decomposition and slag equations, then explain why neither gives iron metal directly.

Give me a hint

Track calcium-containing substances rather than adding iron from nowhere.

Compare my reasoning
  1. CaCO₃→CaO+CO₂ is decomposition.
  2. CaO+SiO₂→CaSiO₃ is slag formation.
  3. Neither equation contains iron; iron is produced in Fe₂O₃ reduction.

CaCO₃→CaO+CO₂; CaO+SiO₂→CaSiO₃; iron comes from the separate oxide-reduction equation.

Look for these in your work

  • All formulas and atoms are correct.
  • The roles stay distinct from iron production.
3 · Transfer

A hypothetical 400 g ore contains 80% Fe₂O₃ by mass. Using Fe=56 and O=16, find theoretical iron assuming complete reduction and sufficient CO, and explain why actual recovery may be smaller.

Give me a hint

Apply purity before the oxide-to-metal mass ratio.

Compare my reasoning
  1. Pure Fe₂O₃ amount is 400×0.80=320 g.
  2. Theoretical iron is 320×112/160=224 g.
  3. Incomplete conversion or product losses can make actual recovered mass smaller than this ideal result.

224 g theoretical iron; actual recovery can be lower through incomplete conversion or losses.

Look for these in your work

  • Ore purity is applied first.
  • Theoretical and actual recovery are distinguished.

Common mistakes

  • Writing CO₂ as the reducing reactant instead of CO in hematite reduction.
  • Changing Fe₂O₃ to FeO to balance the required equation.
  • Using an impure ore's whole mass as pure hematite.
  • Calling every furnace step the same redox equation.
Recall without your notes

What can you explain now?

Balance the required hematite reduction equation using carbon monoxide.

Compare with the explanation

Fe₂O₃+3CO→2Fe+3CO₂.

Two iron atoms and six total oxygen atoms are conserved on each side; CO receives the oxide's oxygen.

After trying it yourself, choose your next review. This is your self-assessment.

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Make it stick

Write all five equations, balance each and label its purpose. Calculate a new pure-oxide or impure-ore theoretical yield with the assumptions stated.

If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.

Keep learning Sciences — Co-ordinated (Double)

Course scope and assessment

This 2028–2029 pathway maps all 152 detailed subject-content sections in the published 0654 syllabus to original teaching and staged practice. It adapts teaching from the related single-science pathways and excludes clear 0610-only topics such as classification systems, human excretion, tropic responses, food supply, population dynamics, detailed pollution and biotechnology. Some explanations may include wider single-science context. A 0654-specific objective-by-objective review, Core/Supplement audit, human subject-teacher review, supervised laboratory experience and cumulative assessment remain in progress. Written practice is self-checked and does not predict an official grade. The 2025–2027 syllabus has not been reconciled here; choose the 2028–2029 syllabus only.

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