QUESTION 5 (10 marks)
A sample of iron ore was tested for its iron content using the experimental procedure outlined.
All the iron (Fe) in the sample was converted to Fe2+(aq) by reacting it with H2SO4(aq), forming hydrogen gas. The solution made up to a final volume of 500.0 mL. A 25.00 mL aliquot of the Fe2+ aqueous solution was titrated with a standardised solution of 0.0500 M KMnO4. An average titre of 16.40 mL was obtained.
a) Write a balanced chemical equation for the reaction between the iron (Fe) in the ore sample and sulfuric acid.
[1 mark]
b) Identify the species oxidised in the reaction in Question 5a). Explain your reasoning.
[2 marks]
c) Apply your understanding of half-equations to balance the redox equation.
[2 marks]
d) Calculate the percentage of iron (Fe) in the ore sample. Show your working.
[5 marks]
QCAA guide · typeset solution
QCAA sample response and mark allocation
5a) | Fe(s) + H2SO4 (aq) →FeSO4(aq) + H2(g) | • correctly identifies the balanced equation [1 mark] | |
5b) | Fe (iron). Fe loses two electrons to form Fe2+(aq). | • correctly identifies that Fe (iron) is oxidised [1 mark] • indicates that Fe loses electrons to form Fe2+ [1 mark] | Acceptable responses are: - iron - Fe. Acceptable response is oxidation number of Fe increases from 0 to +2. |
5c) | −(aq) + 8H+(aq) + 5Fe2+(aq) → Mn2+(aq) + 4H2O(l) + MnO4 5Fe3+(aq) | • correctly balances 8H+ + 4H2O in equation [1 mark] • correctly balances 5Fe2+ + 5Fe3+ in equation [1 mark] | |
5d) | –(aq) reacted = 0.01640 × 0.05 = 8.2 × 10−4 Moles MnO4 – : 5Fe2+ MnO4 Moles of Fe2+ in 25.00 mL = 5 × (8.2 × 10−4) = 4.1 × 10−3 mol Moles of Fe2+ in 500.0 mL = moles Fe in sample = 4.1 × 10−3 × 0.5000 ÷ 0.025 = 0.082 = 8.2 × 10−2 mol Mass Fe dissolved = 0.082 × 55.85 = 4.6 g (4.5797) % Fe in iron ore = 4.6 ÷ 8.00 = 57.2% (57.24625) Percentage Fe in ore sample = 57.2% (to one decimal place) | • correctly determines nMnO4 – = 8.2 × 10−4 [1 mark] • determines nFe2+ = 4.1 × 10−3 [1 mark] • determines nFe = 8.2 × 10−2 [1 mark] • determines mass Fe = 4.6 g [1 mark] • determines Fe = 57.2% [1 mark] | Allow FT error from incorrect mole ratio. Acceptable response is mass Fe = 4.5797 g. Acceptable responses are: - % Fe = 57.2 - % Fe = 57.3 |
QCAA sample response and marking criteria reproduced from the official guide.
Compare your working with the guide above.