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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

a. Calculate the volume, in litres, of SO₂ produced

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Answer

To calculate the volume of SO₂ produced when 250 g of Na₂S₂O₅ reacts, we first determine the number of moles of Na₂S₂O₅:

n(Na2S2O5)=250g190g/mol=1.316moln(Na₂S₂O₅) = \frac{250 g}{190 g/mol} = 1.316 mol

According to the reaction:

Na2S2O5(s)+2HCl(aq)→2NaCl(aq)+H2O(l)+2SO2(g)Na₂S₂O₅(s) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + 2SO₂(g)

1 mole of Na₂S₂O₅ produces 2 moles of SO₂. Therefore:

n(SO2)=2×n(Na2S2O5)=2×1.316=2.632moln(SO₂) = 2 \times n(Na₂S₂O₅) = 2 \times 1.316 = 2.632 mol

Using the ideal gas law at 1.00 atm and 15.0°C (converted to Kelvin):

V=nRT/PV = nRT/P

Where:

  • R = 0.0821 L·atm/(K·mol)
  • T = 15.0 + 273.15 = 288.15 K
  • P = 1.00 atm

Plugging in the values:

V(SO2)=2.632×0.0821×288.151.00V(SO₂) = \frac{2.632 \times 0.0821 \times 288.15}{1.00}

Calculating gives: V(SO2)=62.2LV(SO₂) = 62.2 L

Step 2

b.i. Write an overall balanced chemical equation for the reaction that occurs

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Answer

The overall balanced equation for the reaction between SO₂ and I₃⁻ is:

SO2(aq)+2H2O(l)+I3−(aq)→4H+(aq)+SO42−(aq)+3I−(aq)SO₂(aq) + 2H₂O(l) + I₃⁻(aq) → 4H⁺(aq) + SO₄²⁻(aq) + 3I⁻(aq)

The reductant in this reaction is SO₂.

Step 3

b.ii. 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:

n(I3−)=C×Vn(I₃⁻) = C \times V

Where:

  • C = 0.0125 M (concentration of solution A)
  • V = 0.0500 L (volume of solution A)

Thus:

n(I3−)=0.0125×0.0500=6.25×10−4moln(I₃⁻) = 0.0125 \times 0.0500 = 6.25 \times 10^{-4} mol

Step 4

b.iii. Calculate the original concentration of SO₂ in solution A

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Answer

To determine the original concentration of SO₂ in solution A:

First, calculate the moles of I₃⁻ that reacted with Na₂S₂O₃:

Using 14.70 mL of a 0.00850 M solution:

n(Na2S2O3)=0.00850×0.01470=1.25×10−4moln(Na₂S₂O₃) = 0.00850 \times 0.01470 = 1.25 \times 10^{-4} mol

From the reaction stoichiometry, the moles of I₃⁻ that reacted is:

n(I3−)final=n(I3−)initial−n(I3−)in excessn(I₃⁻)_{final} = n(I₃⁻)_{initial} - n(I₃⁻)_{in\ excess}

Since 6.25 \times 10^{-4} mol was added initially and 1.25 \times 10^{-4} mol reacted:

n(I3−)in excess=6.25×10−4−1.25×10−4=5.00×10−4moln(I₃⁻)_{in\ excess} = 6.25 \times 10^{-4} - 1.25 \times 10^{-4} = 5.00 \times 10^{-4} mol

Therefore, the concentration of SO₂ in solution A can be calculated as:

Taking into account the dilution:

C(SO2)=n(SO2)Vtotal=1.25×10−40.100=5.00×10−3MC(SO₂) = \frac{n(SO₂)}{V_{total}} = \frac{1.25 \times 10^{-4}}{0.100} = 5.00 \times 10^{-3} M

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