Problem 39 - Chemistry and Chemical Reactivity

Question

Calcium oxide, \(\mathrm{CaO}\), is used to remove \(\mathrm{SO}_{2}\) from ower plant exhaust. These two compounds react o give solid \(\mathrm{CaSO}_{3} .\) What mass of \(\mathrm{SO}_{2}\) can be emoved using \(1.2 \times 10^{3} \mathrm{kg}\) of \(\mathrm{CaO} ?\)

Step-by-Step Solution

Answer

1370 kg of \( \mathrm{SO}_{2} \) can be removed using 1200 kg of \( \mathrm{CaO} \).

Step 1: Write the Balanced Chemical Equation

The reaction between calcium oxide \( \mathrm{CaO} \) and sulfur dioxide \( \mathrm{SO}_{2} \) forms calcium sulfite \( \mathrm{CaSO}_{3} \). The balanced chemical equation is: \[ \mathrm{CaO} + \mathrm{SO}_{2} \rightarrow \mathrm{CaSO}_{3} \]

Step 2: Calculate Molar Masses

Calculate the molar masses of \( \mathrm{CaO} \) and \( \mathrm{SO}_{2} \).\[ \mathrm{CaO} = 40.08\, (\text{Ca}) + 16.00\, (\text{O}) = 56.08\, \text{g/mol} \] For \( \mathrm{SO}_{2} \):\[ \mathrm{SO}_{2} = 32.07\, (\text{S}) + 2 \times 16.00\, (\text{O}) = 64.07\, \text{g/mol} \]

Step 3: Determine Moles of \( \mathrm{CaO} \)

Convert the given mass of \( \mathrm{CaO} \) to moles using its molar mass.\[ \text{Moles of } \mathrm{CaO} = \frac{1.2 \times 10^{3}\, \text{kg} \times 1000 \frac{\text{g}}{\text{kg}}}{56.08 \frac{\text{g}}{\text{mol}}} \approx 21400 \text{ moles} \]

Step 4: Relate Moles of \( \mathrm{CaO} \) to \( \mathrm{SO}_{2} \)

The balanced equation shows a 1:1 mole ratio between \( \mathrm{CaO} \) and \( \mathrm{SO}_{2} \). Therefore, the moles of \( \mathrm{SO}_{2} \) are equal to the moles of \( \mathrm{CaO} \). Thus, \( \text{Moles of } \mathrm{SO}_{2} = 21400 \text{ moles} \).

Step 5: Calculate Mass of \( \mathrm{SO}_{2} \) Removed

Convert the moles of \( \mathrm{SO}_{2} \) to mass using its molar mass.\[ \text{Mass of } \mathrm{SO}_{2} = 21400 \text{ moles} \times 64.07 \frac{\text{g}}{\text{mol}} \approx 1.37 \times 10^{6} \text{ g} \] Convert the mass from grams to kilograms:\[ 1.37 \times 10^{6} \text{ g} = 1370 \text{ kg} \]

Key Concepts

Molar Mass Calculation

Molar mass is a key concept in stoichiometry that helps in the calculation of the amount of substance. Every element in the periodic table has an atomic mass. To find the molar mass of a compound, add up the atomic masses of all the atoms in a molecule. For example, to calculate the molar mass of calcium oxide (\( \mathrm{CaO} \)), we add the atomic mass of calcium (40.08 g/mol) and oxygen (16.00 g/mol), resulting in a molar mass of 56.08 g/mol.

For sulfur dioxide (\( \mathrm{SO}_{2} \)), the process is similar: combine the atomic mass of sulfur (32.07 g/mol) and the atomic masses of two oxygen atoms (2 x 16.00 g/mol), resulting in 64.07 g/mol. These calculations are essential for converting between moles and grams, which is often required when solving chemical reaction problems.

Balanced Chemical Equation

A balanced chemical equation is essential for accurately describing a chemical reaction. It ensures that the number of atoms for each element is the same on both the reactant and the product sides of the equation. This is guided by the Law of Conservation of Mass, which dictates that mass cannot be created or destroyed in a chemical reaction.

In this specific exercise, calcium oxide (\( \mathrm{CaO} \)) reacts with sulfur dioxide (\( \mathrm{SO}_{2} \)) to form calcium sulfite (\( \mathrm{CaSO}_{3} \)). The balanced chemical equation is:\[ \mathrm{CaO} + \mathrm{SO}_{2} \rightarrow \mathrm{CaSO}_{3} \]

This balanced equation reflects a 1:1 ratio between \( \mathrm{CaO} \) and \( \mathrm{SO}_{2} \), meaning one mole of \( \mathrm{CaO} \) reacts with one mole of \( \mathrm{SO}_{2} \) to form one mole of \( \mathrm{CaSO}_{3} \).

Mole Ratio

The mole ratio is a concept derived from the coefficients of a balanced chemical equation. It is used to determine the relationship between the amount of reactants and products in a reaction. The balanced equation for this problem\( \mathrm{CaO} + \mathrm{SO}_{2} \rightarrow \mathrm{CaSO}_{3} \) shows a 1:1 mole ratio between the reactants \( \mathrm{CaO} \) and \( \mathrm{SO}_{2} \).

This ratio implies that for every mole of \( \mathrm{CaO} \) used, one mole of \( \mathrm{SO}_{2} \) is also consumed. This mole ratio is fundamental in stoichiometry as it helps to calculate how much of one reactant is needed to react with a given amount of another reactant. In this exercise, knowing the moles of \( \mathrm{CaO} \) allows us to determine the moles of \( \mathrm{SO}_{2} \) that can be removed.

Chemical Reaction

In a chemical reaction, substances called reactants are transformed into different substances known as products. Chemical reactions involve the breaking of old bonds and the formation of new bonds, leading to a change in the chemical identity of the substances involved.

For the exercise, the chemical reaction is between calcium oxide (\( \mathrm{CaO} \)) and sulfur dioxide (\( \mathrm{SO}_{2} \)), resulting in calcium sulfite (\( \mathrm{CaSO}_{3} \)).

• Reactants: \( \mathrm{CaO} \) and \( \mathrm{SO}_{2} \)
• Product: \( \mathrm{CaSO}_{3} \)

This reaction helps in removing \( \mathrm{SO}_{2} \), a harmful gas from power plant exhaust, making it significant in environmental chemistry. Understanding how reactants transform into products helps in controlling and improving such beneficial reactions.