One method proposed for removing \(\mathrm{SO}_{2}\) from the flue gases of power plants involves reaction with aqueous \(\mathrm{H}_{2} \mathrm{~S}\). Elemental sulfur is the product. (a) Write a balanced chemical equation for the reaction. (b) What volume of \(\mathrm{H}_{2} \mathrm{~S}\) at \(27^{\circ} \mathrm{C}\) and 760 torr would be required to remove the \(\mathrm{SO}_{2}\) formed by burning 2.0 tons of coal containing \(3.5 \% \mathrm{~S}\) by mass? (c) What mass of elemental sulfur is produced? Assume that all reactions are \(100 \%\) efficient.

: The reaction between sulfur dioxide (\(\mathrm{SO}_2\)) and hydrogen sulfide (\(\mathrm{H}_{2}\mathrm{S}\)) produces elemental sulfur (\(\mathrm{S}\)) and water (\(\mathrm{H}_{2}\mathrm{O}\)): \[ \mathrm{SO}_2 + 2\mathrm{H}_2\mathrm{S} \rightarrow 3\mathrm{S} + 2\mathrm{H}_2\mathrm{O} \]

: First, we need to determine the moles of sulfur dioxide (\(\mathrm{SO}_2\)) produced by burning 2.0 tons of coal with 3.5% sulfur by mass. The mass of sulfur in the coal is: \[ 2.0 \times 10^6\ \text{g} \times 0.035 = 7.0 \times 10^4\ \text{g} \] Now, let's find the moles of sulfur dioxide produced. We can assume each sulfur atom reacts with oxygen to form \(\mathrm{SO}_2\). The molar mass of sulfur (\(\mathrm{S}\)) is 32.07g/mol, and the molar mass of sulfur dioxide (\(\mathrm{SO}_2\)) is 32.07g/mol + 2(16.00g/mol) = 64.07g/mol. Moles of \(\mathrm{SO}_2\) produced: \[ \frac{7.0 \times 10^4\ \text{g}}{32.07\ \text{g/mol}} \times \frac{1\ \text{mol }\mathrm{SO}_2}{1\ \text{mol }\mathrm{S}} = 2.18 \times 10^3\ \text{mol }\mathrm{SO}_2 \]

: From the balanced chemical equation in step 1, we see that 1 mol of \(\mathrm{SO}_2\) reacts with 2 mol of \(\mathrm{H}_2\mathrm{S}\). Thus, the moles of \(\mathrm{H}_2\mathrm{S}\) required would be double the moles of \(\mathrm{SO}_2\) produced: \[ 2.18 \times 10^3\ \text{mol }\mathrm{SO}_2 \times \frac{2\ \text{mol }\mathrm{H}_2\mathrm{S}}{1\ \text{mol }\mathrm{SO}_2} = 4.36 \times 10^3\ \text{mol }\mathrm{H}_2\mathrm{S} \]

: To find the volume of \(\mathrm{H}_2\mathrm{S}\) required, we can use the ideal gas law, which is given by: \[ PV = nRT \] Where P is the pressure in atm (1 atm = 760 torr), V is the volume in liters, n is the number of moles, R is the gas constant (0.0821 L⋅atm/mol⋅K), and T is the temperature in Kelvin. First, let's convert the temperature from Celsius to Kelvin: \[ T = 27^\circ\mathrm{C} + 273.15 = 300.15\ \mathrm{K} \] Now let's plug the values into the ideal gas law and solve for the volume of \(\mathrm{H}_2\mathrm{S}\). Since the pressure is given in torr, we should convert it to atm: \[ V = \frac{nRT}{P} = \frac{(4.36 \times 10^3\ \text{mol})(0.0821\ \text{L⋅atm/mol⋅K})(300.15\ \mathrm{K})} {760\ \text{torr} \times \frac{1\ \text{atm}}{760\ \text{torr}}} = 1.32 \times 10^5\ \text{L} \] So, the volume of \(\mathrm{H}_2\mathrm{S}\) required is 1.32 x \(10^5\) L.

: From the balanced chemical equation in step 1, we see that 1 mol of \(\mathrm{SO}_2\) reacts to produce 3 mol of elemental sulfur (\(\mathrm{S}\)). Thus, the moles of \(\mathrm{S}\) produced would be triple the moles of \(\mathrm{SO}_2\) produced: \[ 2.18 \times 10^3\ \text{mol }\mathrm{SO}_2 \times \frac{3\ \text{mol }\mathrm{S}}{1\ \text{mol }\mathrm{SO}_2} = 6.54 \times 10^3\ \text{mol }\mathrm{S} \] Finally, let's convert the moles of \(\mathrm{S}\) produced to mass: \[ 6.54 \times 10^3\ \text{mol }\mathrm{S} \times \frac{32.07\ \text{g }\mathrm{S}}{1\ \text{mol }\mathrm{S}} = 2.10 \times 10^5\ \text{g} \] So, the mass of elemental sulfur produced is 2.10 x \(10^5\) g.