An environmental engineer analyzes a sample of air contaminated with sulfur dioxide. To a 500.-mL sample at 700. torr and 38°C, she adds 20.00 mL of 0.01017 M aqueous iodine, which reacts as follows:
SO₂(g) + I₂(aq) + H₂O(l) →
HSO⁻₄(aq) + I⁻(aq) + H(aq)
Excess I₂ reacts with 11.37 mL of 0.0105 M sodium thiosulfate:
I₂(aq) +S₂²⁻O₃(aq) → I⁻(aq) + S₄O²⁻(aq)
What is the volume % of SO₂ in the air sample?

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ArpitaK ArpitaK · Answer 1 · 2022-08-28T19:55:10+02:00

An environmental engineer analyzes a sample of air contaminated with sulfur dioxide. The volume % of SO₂ in the original sample is 1,59%

For the reaction:

2SO₂(g) + I₂(aq) + 2H₂O(l) → 2HSO4¹⁻(aq) + 2I¹⁻(aq) +4H⁺(aq)

The add moles of iodine are: 0,0200L×= 2,034x10⁻⁴ moles of I₂

The moles of thiosulfate for the reaction:

I₂(aq) + 2S₂O₃²⁻(aq) → 2I⁻(aq) + S₄O₆²⁻(aq) are:

0,01137L×= 1,194x10⁻⁴ moles of S₂O₃²⁻

Thus, excess moles of I₂ are: 1,194x10⁻⁴ moles of S₂O₃²⁻×= 5,969x10⁻⁵ moles of I₂

That means that moles of I₂ that react with sulfur dioxide are:

2,034x10⁻⁴ moles of I₂ - 5,969x10⁻⁵ moles of I₂ = 1,4371x10⁻⁴ moles of I₂

These moles of I₂ are: 1,4371x10⁻⁴ moles of I₂× = 2,8742x10⁻⁴ moles of SO₂

Using: V = nRT/P Where n are moles (2,8742x10⁻⁴ moles of SO₂)

R is gas constant (0,082atmL/molK) T is temperature (38°C ≡ 311,15K)

P is pressure (700torr ≡ 0,921 atm)

The volume that SO₂ occupy is:

V = 7,962x10⁻³L ≡ 7,962mL

Thus, volume % is: ×100 = 1.59 Volume% of SO₂ in the air sample

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