Reacting masses: iron from iron(III) oxide

Worked example

Iron(III) oxide is reduced by carbon monoxide: Fe₂O₃ + 3CO → 2Fe + 3CO₂. What mass of iron forms from 348 g of Fe₂O₃? Use M(Fe₂O₃) = 159.6 g mol⁻¹ and M(Fe) = 55.8 g mol⁻¹.

1. 1. Moles from mass: solve for n(Fe₂O₃)

   $n = \dfrac{m}{M}$

   = 2.180 mol

2. 2. Mole ratio from the balanced equation: solve for n(Fe)

   $n_\mathrm{B} = \dfrac{b}{a}\,n_\mathrm{A}$

   = 4.361 mol

3. 3. Moles from mass: solve for m(Fe)

   $m = nM$

   = 243.3 g

Answer: m(Fe) = 243 g

Method

n(Fe₂O₃) = m/M. The equation gives 2 mol Fe per mol Fe₂O₃, so n(Fe) = 2n(Fe₂O₃). Then m(Fe) = n(Fe) × M(Fe).

Common mistakes

• Each Fe₂O₃ gives two Fe atoms: n(Fe) = 2 × n(Fe₂O₃).

Formulas used

• Moles from mass

  $n = \dfrac{m}{M}$

• Mole ratio from the balanced equation

  $n_\mathrm{B} = \dfrac{b}{a}\,n_\mathrm{A}$

More stoichiometry questions

• Percentage yield of an ester
• Atom economy
• Titration: sodium hydroxide against sulfuric acid
• Back titration: purity of calcium carbonate
• Volume of gas from a reacting mass
• Relative molecular mass of a volatile liquid
• Water of crystallisation
• Making and diluting a standard solution
• Number of atoms in a mass of compound
• Empirical formula from the masses of each element
• Oxygen content from combustion analysis
• Iron in a tablet by permanganate titration
• Copper(II) by iodometric titration
• Water hardness by EDTA titration
• Nickel by EDTA back titration