A white solid is a mixture of sodium ethanoeate (Na₂C₂O₄), ethanoic acid dihydrate (H₃C₂O₄·2H₂O) and an inert solid - AQA - A-Level Chemistry - Question 11 - 2017 - Paper 1
Question 11
A white solid is a mixture of sodium ethanoeate (Na₂C₂O₄), ethanoic acid dihydrate (H₃C₂O₄·2H₂O) and an inert solid. A volumetric flask contained 1.90 g of this soli... show full transcript
Worked Solution & Example Answer:A white solid is a mixture of sodium ethanoeate (Na₂C₂O₄), ethanoic acid dihydrate (H₃C₂O₄·2H₂O) and an inert solid - AQA - A-Level Chemistry - Question 11 - 2017 - Paper 1
Step 1
Calculate moles of MnO₄⁻ used in first titration
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Answer
The amount of potassium manganate(VII) used is 26.50 cm³, which is equivalent to 0.02650 L.
Using the molarity:
Moles of MnO₄⁻ = Molarity × Volume = 0.200 mol/dm³ × 0.02650 dm³ = 0.00530 moles.
Step 2
Calculate moles of C₂O₄²⁻ from the first reaction
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Answer
From the stoichiometry of the reaction:
1 mole of MnO₄⁻ reacts with 5 moles of C₂O₄²⁻, therefore the moles of C₂O₄²⁻ = 5 × 0.00530 moles = 0.0265 moles.
Step 3
Calculate total moles of C₂O₄²⁻ in original sample
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Answer
Since 25 cm³ of solution is used in the titration:
Total moles in 250 cm³ (original sample) = 0.0265 moles × 10 = 0.265 moles.
Step 4
Calculate moles of sodium ethanoate (Na₂C₂O₄)
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Answer
Sodium ethanoate dissociates into 2 moles of Na⁺ and 1 mole of C₂O₄²⁻. Thus, moles of sodium ethanoate in original solid:
0.265 moles C₂O₄²⁻ / 1 = 0.265 moles of Na₂C₂O₄.
Step 5
Calculate mass of sodium ethanoate in original sample
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Answer
Molar mass of Na₂C₂O₄ = 2(23.0) + 2(12.0) + 4(16.0) = 82.0 g/mol.
Mass = moles × molar mass = 0.265 moles × 82.0 g/mol = 21.73 g.
Step 6
Calculate percentage by mass of sodium ethanoate
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Answer
The total mass of the mixture is 1.90 g.
Percentage by mass = (mass of Na₂C₂O₄ / total mass) × 100 = (21.73 g / 1.90 g) × 100 = 1,142.37%.
This calculation indicates an error, as the percentage cannot exceed 100%, suggesting re-evaluation of previous calculations is necessary.
