Problem 110 - Chemistry: The Central Science

Question

A particular coal contains \(2.5 \%\) sulfur by mass. When this coal is burned at a power plant, the sulfur is converted into sulfur dioxide gas, which is a pollutant. To reduce sulfur dioxide emissions, calcium oxide (lime) is used. The sulfur dioxide reacts with calcium oxide to form solid calcium sulfite. (a) Write the balanced chemical equation for the reaction. (b) If the coal is burned in a power plant that uses 2000 tons of coal per day, what mass of calcium oxide is required daily to eliminate the sulfur dioxide? (c) How many grams of calcium sulfite are produced daily by this power plant?

Step-by-Step Solution

Answer

(a) The balanced chemical equation for the reaction between sulfur dioxide and calcium oxide is: \(SO2 + CaO → CaSO3\) (b) The required daily mass of calcium oxide to eliminate sulfur dioxide emissions from burning 2000 tons of coal is approximately \(87.55 kg\). (c) The mass of calcium sulfite produced daily by the power plant is approximately \(187.44 kg\).

Step 1: Write the chemical formulas of reactants and products

The reactants in this case are sulfur dioxide (SO2) and calcium oxide (CaO). The product formed is calcium sulfite (CaSO3).

Step 2: Write the unbalanced chemical equation

The unbalanced chemical equation is: SO2 + CaO → CaSO3

Step 3: Balance the chemical equation

The balanced chemical equation for the reaction is: SO2 + CaO → CaSO3 This equation is already balanced. #b. Mass of calcium oxide required daily#

Step 4: Calculate the mass of sulfur in the coal

Given that 2.5% of the coal's mass is sulfur, we first calculate the daily mass of sulfur using: Daily mass of sulfur = (2.5/100) × (2000 tons) Daily mass of sulfur = 50 tons of sulfur

Step 5: Calculate the moles of sulfur

We'll use the molar mass of sulfur to convert the mass to moles: Moles of sulfur = (50 tons) × (1 ton / 1000 kg) × (1000 g / 1 kg) × (1 mol / 32.07 g) Moles of sulfur = 1560.62 moles

Step 6: Calculate the moles of calcium oxide needed

According to the balanced equation, one mole of sulfur dioxide (which comes from one mole of sulfur) needs one mole of calcium oxide to react. Therefore, we need the same amount of moles of calcium oxide as the moles of sulfur: Moles of calcium oxide = 1560.62 moles

Step 7: Calculate the mass of calcium oxide needed daily

To find the mass of calcium oxide needed, we'll use its molar mass: Mass of calcium oxide = (1560.62 moles) × (56.08 g/mol) Mass of calcium oxide = 87545.29 g or 87.55 kg (approximately) So, the required daily mass of calcium oxide is 87.55 kg. #c. Mass of calcium sulfite produced daily#

Step 8: Calculate the moles of calcium sulfite produced

From the balanced equation, one mole of sulfur reacts with one mole of calcium oxide to produce one mole of calcium sulfite. So the number of moles of calcium sulfite produced is equal to the moles of sulfur: Moles of calcium sulfite = 1560.62 moles

Step 9: Calculate the mass of calcium sulfite produced daily

To find the daily mass of calcium sulfite produced, we use its molar mass: Mass of calcium sulfite = (1560.62 moles) × (120.17 g/mol) Mass of calcium sulfite = 187438.29 g or 187.44 kg (approximately) So, the mass of calcium sulfite produced daily by the power plant is 187.44 kg.

Key Concepts

Stoichiometry

At the heart of understanding chemical reactions lies stoichiometry. It is the study of the quantitative relationships between the reactants and products in chemical reactions. It is akin to recipe calculation in cooking where one must use the precise amount of each ingredient to achieve the desired dish. Similarly, in stoichiometry, chemists need to use the exact proportions of substances to ensure that a reaction goes to completion without excess of any reactant.

In environmental chemistry, stoichiometry is crucial when calculating the amount of materials needed to neutralize a pollutant. For example, knowing the exact amount of calcium oxide to react with sulfur dioxide in a coal power plant ensures that the harmful gas is effectively removed from emissions. The example in the exercise showed how percentage by mass is used to derive quantitative information necessary to perform stoichiometric calculations, such as determining how much sulfur is present in coal and then how much lime is needed daily to reduce sulfur emissions.

Pollutant Reduction

Pollutant reduction represents a significant application of environmental chemistry with the objective of minimizing the harmful effects of pollutants on the environment. In the exercise, sulfur dioxide (SO₂), a notorious pollutant resulting from burning coal, is the focus. When released into the atmosphere, it contributes to acid rain, which can harm ecosystems, corrode structures, and affect human health.

To mitigate these consequences, chemicals like calcium oxide are used to neutralize acidic gases. The process involves a chemical reaction where calcium oxide reacts with sulfur dioxide to form a less harmful solid compound, calcium sulfite. This conversion is an excellent demonstration of using stoichiometry in environmental applications—calculating the precise mass of a reagent needed to ensure complete removal of a pollutant. Hence, understanding the concepts of stoichiometry and balanced equations is essential for effective environmental management and pollution control.

Balanced Chemical Equations

Balanced chemical equations are symbolic representations of a chemical reaction, where the number of atoms for each element is the same on both the reactant and product sides. They are fundamental to the study of chemistry because they ensure the law of conservation of mass is adhered to. In simple terms, what you start with in a reaction must equal what you end up with—no atoms are lost or created in the process.

In the context of environmental chemistry, balanced equations provide us with the mole ratio of reactants to products, which is indispensable when trying to figure out how much reagent is required to react with a given amount of pollutant, as shown in the exercise. When you balance the equation for the reaction between sulfur dioxide and calcium oxide, it confirms that the molar ratio is 1:1, enabling the stoichiometric calculation that ensures all sulfur dioxide can be converted to calcium sulfite. This action not only helps in pollutant reduction but also helps in conserving resources by avoiding the use of excess reagents.