Q93 · Practice questionComplex unfamiliar6 marks
QUESTION 93 (6 marks)
A 2.00 mL hydrogen peroxide sample reacts completely with 20.00 mL of 0.05000 mol L⁻¹ Fe²⁺ in excess acid. Peroxide is reduced according to while Fe²⁺ forms Fe³⁺. The remaining Fe²⁺ requires 10.00 mL of 0.01000 mol L⁻¹ permanganate. Each mole of permanganate reacts with five moles Fe²⁺.
a) Write the overall peroxide–iron ionic equation. [2 marks]
b) Calculate the moles of Fe²⁺ remaining and the moles consumed by peroxide. [2 marks]
c) Calculate the peroxide concentration. [2 marks]
WORKED SOLUTION
6 marksPractice marking scheme
; remaining and consumed Fe²⁺ both mol; peroxide 0.1250 mol L⁻¹.
Initial iron moles=0.001000. Permanganate moles=0.0001000, so remaining iron=0.0005000. Difference consumed by peroxide=0.0005000. One peroxide consumes two Fe²⁺, giving 0.0002500 mol peroxide in 0.00200 L.
- Writes correct iron oxidation half-equation/electron matching. [1 marks]
- Writes balanced peroxide–iron net equation. [1 marks]
- Calculates remaining Fe²⁺ = 0.0005000 mol. [1 marks]
- Calculates consumed Fe²⁺ = 0.0005000 mol. [1 marks]
- Uses two Fe²⁺ per peroxide. [1 marks]
- Calculates 0.1250 mol L⁻¹ peroxide. [1 marks]
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