Problem 99 - Chemistry The Central Science In Si Units

Question

A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the air sample through a "bubbler" containing sodium iodide, which removes the ozone according to the following equation: $$ \begin{aligned} \mathrm{O}_{3}(g)+2 \mathrm{NaI}(a q)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \\ \mathrm{O}_{2}(g)+\mathrm{I}_{2}(s)+2 \mathrm{NaOH}(a q) \end{aligned} $$ (a) How many moles of sodium iodide are needed to remove \(5.95 \times 10^{-6} \mathrm{~mol}\) of \(\mathrm{O}_{3} ?(\mathbf{b})\) How many grams of sodium iodide are needed to remove \(1.3 \mathrm{mg}\) of \(\mathrm{O}_{3}\) ?

Step-by-Step Solution

Answer

(a) 1.19 × 10^{-5} moles of NaI; (b) 8.13 mg of NaI.

Step 1: Understand the Reaction Equation

The given chemical reaction is \( \mathrm{O}_{3}(g) + 2 \mathrm{NaI}(a q) + \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{O}_{2}(g) + \mathrm{I}_{2}(s) + 2 \mathrm{NaOH}(a q) \). From the equation, for every mole of \( \mathrm{O}_3 \), two moles of \( \mathrm{NaI} \) are required for the reaction.

Step 2: Calculate Moles of Sodium Iodide for Ozone (Part a)

You have \( 5.95 \times 10^{-6} \) moles of \( \mathrm{O}_3 \). According to the reaction stoichiometry, it takes 2 moles of \( \mathrm{NaI} \) to react with 1 mole of \( \mathrm{O}_3 \). Hence, moles of \( \mathrm{NaI} \) needed = \( 2 \times 5.95 \times 10^{-6} = 1.19 \times 10^{-5} \text{ moles} \).

Step 3: Use Molar Mass to Find Grams of Ozone (Part b)

First convert the mass of ozone to moles using its molar mass. Ozone \( (\mathrm{O}_3) \) has a molar mass of approximately 48 g/mol. Thus, moles of \( \mathrm{O}_3 \) in \( 1.3 \text{ mg} = 0.0013 \text{ g} \) is \( \frac{0.0013}{48} = 2.71 \times 10^{-5} \text{ moles} \).

Step 4: Calculate Moles of Sodium Iodide for Given Ozone (Part b)

Now, using \( 2.71 \times 10^{-5} \) moles of \( \mathrm{O}_3 \), the moles of \( \mathrm{NaI} \) required is \( 2 \times 2.71 \times 10^{-5} = 5.42 \times 10^{-5} \text{ moles} \).

Step 5: Convert Moles of Sodium Iodide to Grams (Part b)

The molar mass of \( \mathrm{NaI} \) is approximately 150 g/mol. Therefore, the mass of \( \mathrm{NaI} \) needed is \( 5.42 \times 10^{-5} \times 150 = 0.00813 \text{ g} \) or \( 8.13 \text{ mg} \).

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that involves calculating the quantities of reactants and products in chemical reactions. When you look at a balanced chemical equation, it’s like a recipe—it tells you the exact amounts of ingredients (reactants) needed to make a certain amount of products. In our exercise, the reaction involves ozone, sodium iodide, and water. Knowing the stoichiometric relationships helps us determine how much sodium iodide is needed based on the amount of ozone present.

In the reaction: \(\mathrm{O}_{3}(g) + 2 \mathrm{NaI}(aq) + \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{O}_{2}(g) + \mathrm{I}_{2}(s) + 2 \mathrm{NaOH}(aq)\), the coefficients (numbers in front of molecules) show the mole ratio between the reactants and products. Specifically, for every mole of ozone, you need two moles of sodium iodide. This direct proportion allows us to calculate exactly how much of one substance is needed based on the known quantity of another.

Understanding stoichiometry is critical because it allows chemists to:

• Predict the amount of products formed in a reaction.
• Determine how much reactant is necessary to completely react with another.
• Scale reactions up or down based on the need, such as in industrial applications.

Chemical Reactions

Chemical reactions describe the process where substances interact to form new compounds. Every reaction can be understood as a transformation where starting materials (reactants) are converted into different materials (products). The ozone concentration measurement reaction involves ozone reacting with sodium iodide to form different products.

In our specific reaction, ozone \(\mathrm{O}_3\) is the primary reactant being measured. It's a gas that reacts with aqueous sodium iodide \(2\mathrm{NaI}(aq)\) in the presence of water \(\mathrm{H}_2\mathrm{O}(l)\). The products are oxygen gas \(\mathrm{O}_2(g)\), solid iodine \(\mathrm{I}_2(s)\), and aqueous sodium hydroxide \(2\mathrm{NaOH}(aq)\). This transformation is significant because it allows the removal of ozone from air samples for measurement.

Each reaction involves breaking and forming chemical bonds. In this case, ozone molecules break apart, allowing iodine ions in sodium iodide to form iodine molecules, and sodium combines with oxygen and hydrogen to form sodium hydroxide. These specific interactions are governed by the principles of conservation of mass and energy, indicating the total amount of each element is constant throughout the reaction.

Mole Calculations

Mole calculations are crucial in chemistry for converting between the number of molecules and the measurable amounts of substances. The mole is a bridge between the atomic scale and the real-world scale since it allows us to count atoms and molecules by weighing them.

In our exercise, we calculate the moles of sodium iodide needed using the known amount of ozone in moles. We use the reaction’s stoichiometry to determine that for each mole of ozone, two moles of sodium iodide react.

To find the amount of sodium iodide needed in grams, we:

• Calculate the moles of ozone present.
• Apply the stoichiometric factor from the balanced equation (2 moles of \(\mathrm{NaI}\) per mole of \(\mathrm{O}_3\)).
• Convert the resulting moles of \(\mathrm{NaI}\) to grams using the molecular weight of sodium iodide (approximately 150 g/mol).

This involves both conversions between mass and moles, as well as understanding and applying stoichiometric ratios. By performing mole calculations, you gain precision and accuracy in chemical analysis, as demonstrated in determining how much sodium iodide is necessary to remove a specific amount of ozone.