Question

Sulfur dioxide can be removed from the smokestack emissions of power plants by reacting it with hydrogen sulfide, producing sulfur and water. What volume of hydrogen sulfide at \(27^{\circ} \mathrm{C}\) and \(755 \mathrm{~mm} \mathrm{Hg}\) is required to remove the sulfur dioxide produced by a power plant that burns one metric ton of coal containing \(5.0 \%\) sulfur by mass? How many grams of sulfur are produced by the reaction of \(\mathrm{H}_{2} \mathrm{~S}\) with \(\mathrm{SO}_{2} ?\)

Short answer

Answer: 25,682.33 L of hydrogen sulfide is required to remove the sulfur dioxide produced, and the mass of sulfur produced in the reaction is 50 kg.

Step-by-step solution

Determine the mass of sulfur in 1 metric ton of coal

We are given that the coal contains 5.0% sulfur by mass. So, first find the mass of sulfur present in 1 metric ton (1000 kg) of coal: Sulfur mass = (mass of coal) * (percentage of sulfur) Sulfur mass = 1000 kg * (5.0/100) = 50 kg

Write the balanced chemical equation for the reaction

The chemical reaction between hydrogen sulfide (H2S) and sulfur dioxide (SO2) producing sulfur (S) and water (H2O) is as follows: 2 H2S + SO2 -> 3 S + 2 H2O Make sure that the equation is balanced, with the same number of atoms on both sides.

Convert the mass of sulfur to moles

To find the volume of H2S required, we need to know the moles of sulfur produced. To do this, divide the mass of sulfur by its molar mass. Molar mass of S = 32.07 g/mol Mass of sulfur = 50 kg * (1000 g/kg) = 50,000 g Moles of sulfur = (mass of sulfur) / (molar mass of S) Moles of sulfur = 50,000 g / 32.07 g/mol = 1,558.05 mol

Determine the moles of H2S needed

From the balanced chemical equation, we see that 2 moles of H2S react with 1 mole of SO2 to produce 3 moles of sulfur. Therefore: Moles of H2S = (2 / 3) * moles of sulfur Moles of H2S = (2 / 3) * 1,558.05 mol = 1,038.7 mol

Use the ideal gas law to find the volume of H2S

Apply the ideal gas law, PV = nRT, to find the volume of H2S, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature: V = (nRT) / P Given, temperature T = \(27^{\circ} \mathrm{C} = 300.15 \mathrm{K}\) Pressure P = \(755 \mathrm{~mm} \mathrm{Hg} = 755 / 760 \mathrm{atm} = 0.9934 \mathrm{atm}\) R (Ideal gas constant) = 0.0821 \(\mathrm{L \cdot atm / (mol \cdot K)}\) V = (1,038.7 mol * 0.0821 \(\mathrm{L \cdot atm / (mol \cdot K)}\) * 300.15 K) / 0.9934 atm = 25,682.33 L So, the volume of H2S required to remove the sulfur dioxide produced is 25,682.33 L

Calculate the mass of sulfur produced in the reaction

The mass of sulfur produced in the reaction equals the mass of sulfur present in 1 metric ton of coal. So, the mass of sulfur produced is 50 kg. Therefore, 25,682.33 L of hydrogen sulfide is required to remove the sulfur dioxide produced by burning one metric ton of coal containing 5.0% sulfur by mass, and 50 kg of sulfur is produced in the reaction.