Metals below carbon in the reactivity series

2 marks · Easy

Given the equation: \ce{Fe2O3(s) + 3CO(g) -> 2Fe(l) + 3CO2(g)}. Calculate the mass of iron produced when 32.0 g of Fe2O3 reacts completely. (Relative atomic masses: Fe = 56, O = 16)

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22.4 g (±0.2)

Common mistakes: Only 1 Fe calculated instead of 2; Using incorrect molar mass

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Marking points

M1 · 1 mark · Correctly calculates moles of Fe2O3 and uses 2:1 molar ratio to Fe. (follow-through)

A1 · 1 mark · Gives mass of iron produced as 22.4 g (allow rounding within tolerance). (follow-through)

Worked solution

Step 1: Molar mass of Fe2O3 = (2 × 56) + (3 × 16) = 112 + 48 = 160 g/mol Moles of Fe2O3 = 32.0 g / 160 g/mol = 0.20 mol Step 2: From the equation, 1 mol Fe2O3 produces 2 mol Fe. Moles of Fe = 0.20 × 2 = 0.40 mol Mass of Fe = 0.40 mol × 56 g/mol = 22.4 g