Problem 79 - Chemistry The Central Science

Question

In 1986 an electrical power plant in Taylorsville, Georgia, burned \(8,376,726\) tons of coal, a national record at that time. (a) Assuming that the coal was 83\(\%\) carbon and 2.5\(\%\) sulfur and that combustion was complete, calculate the number of tons of carbon dioxide and sulfur dioxide produced by the plant during the year. (b) If 55\(\%\) of the SO \(_{2}\) could be removed by reaction with powdered CaO to form \(\mathrm{CaSO}_{3},\) how many tons of \(\mathrm{CaSO}_{3}\) would be produced?

Step-by-Step Solution

Answer

In 1986, the electrical power plant produced 25,597,760.06 tons of CO₂ and 418,836.3 tons of SO₂. If 55% of the SO₂ could be removed by reaction with powdered CaO, 431,174.93875 tons of CaSO₃ would be produced.

Step 1: Calculate tons of carbon and sulfur in coal

To calculate the mass of carbon and sulfur in the coal, we will first convert the percentage to a decimal and then multiply it by the total mass of the coal. Mass of carbon = (Percentage of carbon in coal) × (Total mass of coal) Mass of carbon = (0.83) × (8,376,726 tons) = 6,952,582.58 tons Mass of sulfur = (Percentage of sulfur in coal) × (Total mass of coal) Mass of sulfur = (0.025) × (8,376,726 tons) = 209,418.15 tons

Step 2: Calculate tons of CO₂ and SO₂ produced

Complete combustion means that all carbon will convert to CO₂ and all sulfur will convert to SO₂. We will use the stoichiometry to calculate the mass of CO₂ and SO₂ produced. The balanced chemical equation for combustion of carbon and sulfur is as follows: C + O₂ → CO₂ S + O₂ → SO₂ Molar mass of C = 12 g/mol, Molar mass of CO₂ = 44 g/mol Molar mass of S = 32 g/mol, Molar mass of SO₂ = 64 g/mol Mass of CO₂ = (Mass of carbon) × (Molar mass of CO₂ / Molar mass of C) Mass of CO₂ = (6,952,582.58 tons) × (44/12) = 25,597,760.06 tons Mass of SO₂ = (Mass of sulfur) × (Molar mass of SO₂ / Molar mass of S) Mass of SO₂ = (209,418.15 tons) × (64/32) = 418,836.3 tons

Step 3: Calculate the amount of SO₂ removed by reaction with CaO

To find the amount of SO₂ removed by reaction with CaO, we will first find the 55% of the total mass of SO₂ produced, and then use stoichiometry to calculate the mass of CaSO₃ formed. The balanced chemical equation for the reaction is as follows: SO₂ + CaO → CaSO₃ Mass of SO₂ removed = (Percentage of SO₂ removed) × (Total mass of SO₂ produced) Mass of SO₂ removed = (0.55) × (418,836.3 tons) = 230,359.965 tons

Step 4: Calculate the mass of CaSO₃ produced

Now, we will use stoichiometry to calculate the mass of CaSO₃ produced from the reaction between SO₂ and CaO. Molar mass of CaO = 56 g/mol, Molar mass of CaSO₃ = 120 g/mol Mass of CaSO₃ produced = (Mass of SO₂ removed) × (Molar mass of CaSO₃ / Molar mass of SO₂) Mass of CaSO₃ produced = (230,359.965 tons) × (120/64) = 431,174.93875 tons The answers are: (a) 25,597,760.06 tons of CO₂, and 418,836.3 tons of SO₂ were produced. (b) 431,174.93875 tons of CaSO₃ would be produced if 55% of the SO₂ could be removed by reaction with powdered CaO.

Key Concepts

Combustion Reactions

Combustion reactions are chemical reactions that occur when a substance combines with oxygen, releasing energy. In our context, this involves burning coal, primarily composed of carbon and sulfur.
During the combustion of coal, carbon reacts with oxygen in the air to produce carbon dioxide \( ( \text{C} + \text{O}_2 \rightarrow \text{CO}_2) \). Likewise, sulfur reacts with oxygen to form sulfur dioxide \( ( \text{S} + \text{O}_2 \rightarrow \text{SO}_2) \).
The energy released in these reactions is what powers electrical plants. However, these reactions also produce gases that contribute to pollution, like carbon dioxide (a greenhouse gas) and sulfur dioxide (which can cause acid rain). Understanding full combustion ensures we can calculate pollutant production and take measures to control it.

Chemical Equations

Chemical equations are written representations of chemical reactions. They show reactants transforming into products, maintaining the conservation of mass and atoms. In this exercise, the chemical equations we used were:

• \( \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \)
• \( \text{S} + \text{O}_2 \rightarrow \text{SO}_2 \)
• \( \text{SO}_2 + \text{CaO} \rightarrow \text{CaSO}_3 \)

For complete combustion, all carbon and sulfur in the coal must be accounted for, converting fully into their respective oxides.
This exercise also shows a reaction between sulfur dioxide and calcium oxide, a crucial step for pollution control. Each equation illustrates the inputs and outputs of the reactions involved, helping us calculate the mass of pollutants and byproducts formed.

Molar Mass

Molar mass is a measure of the mass of one mole of a substance, expressed in grams per mole \((\text{g/mol})\). It serves as a bridge to relate amounts of substances involved in chemical reactions.
In the solution, we used the molar masses of various compounds:

• Carbon (\( \text{C} \)): 12 \( \text{g/mol} \)
• Carbon dioxide (\( \text{CO}_2 \)): 44 \( \text{g/mol} \)
• Sulfur (\( \text{S} \)): 32 \( \text{g/mol} \)
• Sulfur dioxide (\( \text{SO}_2 \)): 64 \( \text{g/mol} \)
• Calcium oxide (\( \text{CaO} \)): 56 \( \text{g/mol} \)
• Calcium sulfite (\( \text{CaSO}_3 \)): 120 \( \text{g/mol} \)

These values are crucial for using stoichiometry, allowing us to convert from moles to grams and vice versa. This conversion is essential for calculating how much carbon dioxide and sulfur dioxide are produced and the amount of calcium sulfite formed.

Pollution Control

Pollution control is the process of managing and reducing pollutants released into the environment. In this scenario, it primarily involves capturing sulfur dioxide (\(\text{SO}_2\)) produced during coal combustion.
Sulfur dioxide is an air pollutant that can lead to acid rain, which negatively affects ecosystems and structures. To minimize its release, power plants utilize reactions like \(\text{SO}_2 + \text{CaO} \rightarrow \text{CaSO}_3\).
Here, calcium oxide reacts with sulfur dioxide to form calcium sulfite, which can be safely disposed of or converted into useful products. This step was crucial because 55% of \(\text{SO}_2\) was removed, greatly reducing potential environmental harm. Understanding stoichiometry in pollution control helps estimate the effectiveness of these measures, ensuring cleaner air and compliance with environmental regulations.