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Sulfur dioxide gas is commonly used as a preservative in wine - VCE - SSCE Chemistry - Question 7 - 2007 - Paper 1

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Sulfur dioxide gas is commonly used as a preservative in wine. An important source of SO₂ is solid sodium metabisulfite (Na₂S₂O₅; molar mass 190 g mol⁻¹). Na₂S₂O₅ re... show full transcript

Worked Solution & Example Answer:Sulfur dioxide gas is commonly used as a preservative in wine - VCE - SSCE Chemistry - Question 7 - 2007 - Paper 1

Step 1

Calculate the volume, in litres, of SO₂ produced

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Answer

To find the volume of SO₂ produced at 1.00 atm and 15.0°C when 250 g of Na₂S₂O₅ reacts:

  1. Calculate the moles of Na₂S₂O₅: n(Na2S2O5)=250g190g/mol=1.3158moln(Na₂S₂O₅) = \frac{250 g}{190 g/mol} = 1.3158 mol

  2. Use the stoichiometry from the reaction:

    n(SO2)=2×n(Na2S2O5)=2×1.3158=2.6316moln(SO₂) = 2 \times n(Na₂S₂O₅) = 2 \times 1.3158 = 2.6316 mol

  3. Use the Ideal Gas Law to find volume:

    V = \frac{nRT}{P}\$$ where R = 0.0821 L·atm/(K·mol), T = 15.0 + 273.15 = 288.15 K, $$V = \frac{(2.6316 mol)(0.0821 L·atm/(K·mol))(288.15 K)}{1.00 atm} = 62.2 L$$

Step 2

Write an overall balanced chemical equation for the reaction

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Answer

The overall balanced equation for the reaction is:

SO2(aq)+2H2O(l)⇌4H+(aq)+SO42−(aq)+2e−SO₂(aq) + 2H₂O(l) ⇌ 4H⁺(aq) + SO₄²⁻(aq) + 2e⁻

The reductant in this reaction is SO₂.

Step 3

Calculate the amount, in mol, of I₃⁻ added to solution A

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Answer

To calculate the amount of I₃⁻ added to solution A:

  1. Calculate the moles of I₃⁻:

    n(I3−)=0.0125 M×0.0500 L=6.25×10−5 moln(I₃⁻) = 0.0125 \text{ M} \times 0.0500 \text{ L} = 6.25 \times 10^{-5} \text{ mol}

Step 4

Calculate the original concentration of SO₂ in solution A

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Answer

To find the original concentration of SO₂:

  1. Calculate moles of Na₂S₂O₃ reacting with I₃⁻:

    n(Na2S2O3)=0.00850 M×0.01470 L=1.25×10−5 moln(Na₂S₂O₃) = 0.00850 \text{ M} \times 0.01470 \text{ L} = 1.25 \times 10^{-5} \text{ mol}

  2. From the stoichiometry of the reaction:

    n(I3−) reacting with SO2=6.25×10−5−1.25×10−5=5.00×10−5 moln(I₃⁻) \text{ reacting with } SO₂ = 6.25 \times 10^{-5} - 1.25 \times 10^{-5} = 5.00 \times 10^{-5} \text{ mol}

  3. Use the total volume of solution B (0.100 L) to find the concentration of SO₂:

    C(SO2)=5.00×10−5 mol0.100 L=5.00×10−4 MC(SO₂) = \frac{5.00 \times 10^{-5} \text{ mol}}{0.100 \text{ L}} = 5.00 \times 10^{-4} \text{ M}

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