Use the References to access important values if needed for this question. A typ

Question

Use the References to access important values if needed for this 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 s oxidized as the coal burns: 4FeS2(s) + 11 02g)2 Fe203(s) + 8 S02(g) Once in the atmosphere, the SO2 is oxidized to SO3. which then reacts with water in the atmosphere to form sulfuric acid: SO3(8) + H20()>H2SO4(a) If 41.2 metric tons of coal that contains 3.97 % by weight s is burned and all of the sulfuric acid that is formed rains down into a pond of dimensions 393 m x 251 m x 3.94 m, what is the pH of the pond? (OK to assume 2 mol H3O per mole H2SO4) Hint:1 metric ton 1000 kg

Explanation / Answer

The coal contains 3.97% by weight of S; therefore, 100 g of coal contains 3.97 g of S.

Therefore, weight of coal in 41.2 metric tons of coal = (41.2 metric tons coal)*(3.97 g S/100 g coal) = (41.2 metric tons coal)*(1000 kg/1 metric ton)*(1000 g/1 kg)*(3.97 g S/100 g coal) = 1635640 g.

All the S in coal is present as FeS2. Therefore, 1635640 g S is present in 41.2 metric tons coal as FeS2.

Determine the moles of FeS2 that will contain 1635640 g S as FeS2. We start with the molar mass of FeS2. The atomic masses are

Fe: 55.845 u

S: 32.065 u

The gram molar mass of FeS2 = (1*55.845 + 2*32.065) g/mol = 119.975 g/mol.

Moles of FeS2 corresponding to 1635640 g S = (mass of S)/(molar mass of FeS2) = (1635640 g)/(119.975 g/mol) = 13633.1734 mole (keep guard digits extra).

Now set up the equations for the reactions as below.

4 FeS2 (s) + 11 O2 (g) -------> 2 Fe2O3 (s) + 8 SO2 (g) ……(1)

2 SO2 (g) + O2 (g) --------> 2 SO3 (g) …….(2)

SO3 (g) + H2O (l) --------> H2SO4 (aq) ……(3)

H2SO4 (aq) + H2O (l) --------> 2 H3O+ (aq) + SO42- (aq) ……(4)

As per stoichiometric equation (1),

4 moles FeS2 = 8 moles SO2.

Therefore, 13633.1734 moles FeS2 = (13633.1734 mole FeS2)*(8 moles SO2/4 moles FeS2) = 27266.3468 moles SO2.

As per stoichiometric equation (2),

2 moles SO2 = 2 moles SO3.

Therefore, 27266.3468 moles SO2 = (27266.3468 moles SO2)*(2 moles SO3/2 moles SO2) = 27266.3468 moles SO3.

Again, as per stoichiometric equation (3),

1 mole SO3 = 1 mole H2SO4.

Therefore, 27266.3468 moles SO3 = (27266.3468 moles SO3)*(1 mole H2SO4/1 mole SO3) = 27266.3468 moles H2SO4.

Finally, as per stoichiometric equation (4),

1 mole H2SO4 = 2 moles H3O+.

Therefore, 27266.3468 moles H2SO4 = (27266.3468 moles H2SO4)*(2 moles H3O+/1 mole H2SO4) = 54532.6936 moles H3O+.

The dimensions of the pond are 393 m x 251 m x 3.94 m; therefore, the volume of water in the pond = (393*251*3.94) m3 = 388653.42 m3 = (388653.42 m3)*(1000 L/1 m3) = 388653420 L (1 m3 = 1000 L).

The H3O+ ends up in the pond. Therefore, the molarity of H3O+ = (moles of H3O+)/(volume of water in the pond) = (54532.6936 moles)/(388653420 L) = 1.4031*10-4 mol/L.

Assume that the pH of the water in the pond is only due to addition of H3O+. Thus, the pH of the pond is pH = -log [H3O+] where the [..] denote concentration in molarity. Therefore,

pH = -log (1.4031*10-4 mol/L) = 3.8529 ? 3.85 (correct to three sig. figs, ans).

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