Question

An environmental engineer analyzes a sample of air contaminated with sulfur dioxide. To a 500m mL sample of air at 700 torr and 38 degrees celsius, she adds 20.00 mL of 0.01017 M aqueous iodine, which reacts as follows:
2SO2 (g) + I2 (aq) +2H20 (l) ______> 2HSO41 -(aq) +2I1 -(aq) +4H+(aq)

The excess 12 (aq) left over from the above reaction with tirade with 11.37 mL of 0.0105 M thiosulfate (S2O32-) as follows: I2 (aq) + 2S2O32 -(aq) ---->2I1-(aq) +S4O62-(aq). What is the volume % of SO2 in the original sample.

Answer 1

Answer:

The volume % of SO₂ in the original sample is 1,59%

Explanation:

For the reaction:

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

The add moles of iodine are:

0,0200L×$\frac{0,01017mol}{L}$= 2,034x10⁻⁴ moles of I₂

The moles of thiosulfate for the reaction:

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

are:

0,01137L×$\frac{0,0105mol}{L}$= 1,194x10⁻⁴ moles of S₂O₃²⁻

Thus, excess moles of I₂ are:

1,194x10⁻⁴ moles of S₂O₃²⁻×$\frac{1molI_{2}}{2molS_{2}O_{3}^{2-}}$= 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× $\frac{2molSO_2}{1molI_{2}}$ = 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:

$\frac{7,962mLSO_{2}}{500mLair}$×100 = 1,59 Volume%

I hope it helps!