Problem 110 - Chemistry: An Atoms-Focused Approach

Question

Some indoor air-purification systems work by converting a little of the oxygen in the air to ozone, which kills mold and mildew spores and other biological air pollutants. The chemical equation for the ozone generation reaction is $$3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{O}_{3}(g)$$ It is claimed that one such system generates \(4.0 \mathrm{g} \mathrm{O}_{3}\) per hour from dry air passing through the purifier at a flow of \(5.0 \mathrm{L} / \mathrm{min.}\) If exactly 1 liter of indoor air contains \(0.28 \mathrm{g}\) \(\mathrm{O}_{2},\) what percentage of the \(\mathrm{O}_{2}\) is converted to \(\mathrm{O}_{3}\) by the air purifier?

Step-by-Step Solution

Answer

Answer: The air purifier converts approximately 4.76% of the oxygen to ozone.

Step 1: Write down given information

We are given: - The chemical equation for the ozone generation reaction: \(3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{O}_{3}(g)\) - The air purifier generates \(4.0 \mathrm{g} \mathrm{O}_{3}\) per hour. - Air flows through the purifier at a rate of \(5.0 \mathrm{L} / \mathrm{min}\) - 1 liter of indoor air contains \(0.28 \mathrm{g} \mathrm{O}_{2}\)

Step 2: Calculate moles of O₃ generated

Using the molar mass of O₃ (\(48.0 \mathrm{g/mol}\)), we can calculate the moles of O₃ generated per hour: $$ \mathrm{moles \ of \ O}_{3} = \frac{4.0 \mathrm{g} \ \mathrm{O}_{3}}{48.0 \mathrm{g/mol}} = 0.0833 \ \mathrm{mol \ O}_{3} $$

Step 3: Calculate moles of O₂ consumed

From the balanced chemical equation, we can see that 3 moles of O₂ are required to produce 2 moles of O₃. We can use this stoichiometric ratio to calculate the moles of O₂ consumed. $$ \mathrm{moles \ of \ O}_{2} = \frac{3 \ \mathrm{moles \ O}_{2}}{2 \ \mathrm{moles \ O}_{3}} \times 0.0833 \ \mathrm{moles \ O}_{3} = 0.125 \ \mathrm{moles \ O}_{2} $$

Step 4: Convert moles of O₂ consumed to mass

Using the molar mass of O₂ (\(32.0 \mathrm{g/mol}\)), we can convert the moles of O₂ consumed to mass: $$ \mathrm{mass \ of \ O}_{2} = 0.125 \ \mathrm{moles \ O}_{2} \times 32.0 \mathrm{g/mol} = 4.0 \ \mathrm{g \ O}_{2} $$

Step 5: Calculate total mass of O₂ in the air passed through the purifier

Air flows through the purifier at a rate of \(5.0 \mathrm{L/min}\). In an hour, the volume of air passed through the purifier is: $$ 5.0 \ \mathrm{L/min} \times 60 \ \mathrm{min/h} = 300 \ \mathrm{L} $$ Since 1 L of air contains \(0.28 \ \mathrm{g}\) \(\mathrm{O}_{2}\), the total mass of O₂ in the air passed through the purifier is: $$ \mathrm{Total \ mass \ of \ O}_{2} = 300 \ \mathrm{L} \times 0.28 \ \mathrm{g \ O}_{2}/\mathrm{L} = 84 \ \mathrm{g \ O}_{2} $$

Step 6: Calculate the percentage of O₂ converted to O₃

Now, we can find the percentage of O₂ that is converted to O₃ by comparing the mass of O₂ used to produce O₃ (4.0 g) to the total mass of O₂ in the air passed through the purifier (84 g). $$ \mathrm{Percentage \ of \ O}_{2} \ \mathrm{converted \ to \ O}_{3} = \frac{4.0 \ \mathrm{g \ O}_{2}}{84 \ \mathrm{g \ O}_{2}} \times 100\% = 4.76 \% $$ The air purifier converts approximately 4.76% of the oxygen to ozone.

Key Concepts

Ozone Generation

Ozone, or O extsubscript{3}, is a molecule composed of three oxygen atoms. It plays a crucial role in air purification systems. These systems work by converting a portion of oxygen gas (O extsubscript{2}) found in the air into ozone. This conversion occurs through a chemical reaction.
The presence of ozone can help in eliminating biological contaminants such as mold and mildew spores. This is due to its strong oxidizing properties. While beneficial, excessive ozone can be harmful, which is why air purification needs to be controlled.
In the chemical reaction used in this exercise, ozone is generated from oxygen with the equation \(3 ext{O}_{2}(g) \rightarrow 2 ext{O}_{3}(g)\). Here, three molecules of O extsubscript{2} effectively lead to the formation of two molecules of O extsubscript{3}. Understanding this conversion is key to analyzing the efficiency and safety of air purification systems.

Chemical Equation

Chemical equations are vital as they symbolize the reactions that happen between substances. In the context of ozone generation, the equation \(3 ext{O}_{2}(g) \rightarrow 2 ext{O}_{3}(g)\) illustrates the transformation of oxygen into ozone.
An equation consists of reactants on the left and products on the right, separated by an arrow. The equation must be balanced to reflect the law of conservation of mass, ensuring that the number of atoms of each element is the same on both sides.
In this equation:

• The reactant is oxygen gas (O extsubscript{2}) with three molecules involved.
• The product is ozone (O extsubscript{3}), formed with two molecules.

Balancing the molecule counts through coefficients ensures the reaction conforms to physical laws. Understanding chemical equations allows us to deduce the ratio of reactants and products, integral for stoichiometric calculations.

Stoichiometry

Stoichiometry involves using the relationships embedded in chemical equations to calculate quantities of reactants or products. For the given reaction \(3 ext{O}_{2} \rightarrow 2 ext{O}_{3}\), stoichiometry helps determine how much ozone is generated from a known amount of oxygen.
In this example, stoichiometric coefficients (3 for O extsubscript{2} and 2 for O extsubscript{3}) indicate that every 3 moles of oxygen gas produce 2 moles of ozone.
By leveraging this ratio, we can calculate, for instance, the amount of oxygen consumed to produce a certain amount of ozone. This calculation is essential for assessing the efficacy and consumption rates in air purification systems. Stoichiometry not only helps with theoretical predictions but also aids in making practical decisions in chemical engineering and environmental studies.

Molar Mass Calculation

The concept of molar mass is foundational in converting between grams and moles, units often used in chemistry. The molar mass of a substance is the mass of one mole of that substance, typically expressed in grams per mole (g/mol).
In the example exercise, the molar mass of ozone (O extsubscript{3}) is 48.0 g/mol and that of oxygen (O extsubscript{2}) is 32.0 g/mol. This means one mole of ozone weighs 48.0 grams, while one mole of oxygen weighs 32.0 grams.
To find out how many moles correspond to a given mass, you divide the mass by the molar mass. For example, with 4.0 grams of O extsubscript{3}, dividing by 48.0 g/mol yields approximately 0.0833 moles. This conversion is essential in translating chemical equations into measurable quantities that can be practically applied and verified in laboratory or industrial settings.