Question

A typical coal-fired electric generating plant will burn about 3 metric tons of coal per hour.

Most of the coal burned in the United States contains 1 to 4 % by weight sulfur in the form of pyrite, which is oxidized as the coal burns:

4FeS₂(s) + 11 O₂(g) = 2 Fe₂O₃(s) + 8 SO₂(g)

Once in the atmosphere, the SO₂ is oxidized to SO₃, which then reacts with water in the atmosphere to form sulfuric acid:

SO₃(g) + H₂O(l) = H₂SO₄(aq)

If 61.8 metric tons of coal that contains 2.75 % by weight S is burned and all of the sulfuric acid that is formed rains down into a pond of dimensions 328 m x 200 m x 4.22 m, what is the pH of the pond? (OK to assume 2 mol H₃O⁺ per mole H₂SO₄.)

Hint: 1 metric ton = 1000 kg

Answer 1

Sol :-

The equation of oxidation of sulphide ore is given as :

4cFeS₂(s) + 11 O₂ (g) --------------> 2 Fe₂O₃ (s) + 8 SO₂ (g)

The oxidation of sulphur dioxide to trioxide is :

SO₂ + 1/2 O₂ ----> SO₃

The reaction of formation of sulphuric acid from SO₃ is :

SO₃ (g) + H₂O(l) -------------> H₂SO₄ (aq)

Given coal sample weighs 61.8 tons

one ton = 1000 kg

So,

61.8 tons = 61800 Kg

2.75 % by weight S, therefore the mass of sulfur in coal = 2.75 % X 61800 Kg = 1699.5 Kg

Moles of sulfur = mass of sulfur / molar mass = 1699.5 X 1000 g / 32 = 53109.375 moles

This much moles of S will give same moles of SO₂ and hence same moles of SO₃ and hence same moles of H₂SO₄ = 53109.375 moles

Volume of pond = 328 m x 200 m x 4.22 m = 276832 m³ = 276832 X 10³ L

Molarity of H₂SO₄ = moles / volume = 53109.375 moles / 276832 X 10³ L = 1.92 X 10^-3 M

Each moles of H₂SO₄ will give two moles of H+

[H⁺] = 2 X 1.92 X 10^-3 M = 3.84 X 10^-3 M

pH = -log [H+]

pH = - log (3.84 x 10^-3)

pH = 2.42