Problem 93 - Chemistry: The Central Science

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: $$ \mathrm{O}_{3}(g)+2 \mathrm{Nal}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow_{2}(g)+\mathrm{I}_{2}(s)+2 \mathrm{NaOH}(a q) $$ (a) How many moles of sodium iodide are needed to remove \(5.95 \times 10^{-6} \mathrm{~mol}\) of \(\mathrm{O}_{3} ?\) (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) The moles of sodium iodide needed to remove \(5.95 \times 10^{-6} mol\) of ozone is \(1.19 \times 10^{-5} mol\). (b) The mass of sodium iodide needed to remove \(1.3 mg\) of ozone is \(8.123 \times 10^{-3} g\).

Step 1: Write down the balanced chemical equation

The balanced chemical equation is given as: \[\mathrm{O}_{3}(g)+2 \mathrm{Nal}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{O}_{2}(g)+\mathrm{I}_{2}(s)+2 \mathrm{NaOH}(a q)\]

Step 2: Find the stoichiometric ratio between ozone and sodium iodide

From the balanced chemical equation, we can see that 1 mole of ozone reacts with 2 moles of sodium iodide. Thus, the stoichiometric ratio between ozone and sodium iodide is 1:2.

Step 3: Calculate moles of sodium iodide needed to remove given moles of ozone

Now, use the stoichiometric ratio to calculate the moles of sodium iodide needed to remove \(5.95 \times 10^{-6} mol\) of ozone: Moles of sodium iodide = (moles of ozone) × (ratio of sodium iodide to ozone) \(= 5.95 \times 10^{-6} mol \times 2 \) \(= 1.19 \times 10^{-5} mol\) (a) The moles of sodium iodide needed to remove \(5.95 \times 10^{-6} mol\) of ozone is \(1.19 \times 10^{-5} mol\).

Step 4: Calculate moles of ozone for the given mass

Now, we need to find the moles of ozone for a mass of \(1.3 mg\). 1. Convert the mass of ozone to grams: \(1.3 mg = 1.3 × 10^{-3} g\) 2. For ozone, \(\mathrm{O}_{3}\), the molar mass is \(3 \times 16.00 g/mol = 48.00 g/mol\). 3. Moles of ozone = (mass of ozone)/(molar mass of ozone) Moles of ozone = \(\frac{1.3 \times 10^{-3} g}{48.00 g/mol}\) = \(2.708 \times 10^{-5} mol\)

Step 5: Calculate moles and mass of sodium required to remove the calculated moles of ozone

Now, using the stoichiometric ratio, calculate moles of sodium iodide needed to remove \(2.708 \times 10^{-5} mol\) of ozone: Moles of sodium iodide = (moles of ozone) × (ratio of sodium iodide to ozone) \(= 2.708 \times 10^{-5} mol \times 2\) \(= 5.416 \times 10^{-5} mol\) Now, let's calculate the mass of sodium iodide corresponding to the number of moles. For sodium iodide, \(\mathrm{NaI}\), the molar mass is \(22.99 g/mol + 126.90 g/mol = 149.89 g/mol\). Mass of sodium iodide = (moles of sodium iodide) × (molar mass of sodium iodide) \(= (5.416 \times 10^{-5} mol) \times (149.89 g/mol) = 8.123 \times 10^{-3} g\) (b) The mass of sodium iodide needed to remove \(1.3 mg\) of ozone is \(8.123 \times 10^{-3} g\).

Key Concepts

Chemical Reaction Equation

Understanding a chemical reaction equation is crucial when venturing into stoichiometric calculations. To put it simply, these equations are like recipes for chemists—they tell us what reactants are needed and what products will be formed. The equation given, \[\mathrm{O}_{3}(g)+2 \mathrm{Nal}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{O}_{2}(g)+\mathrm{I}_{2}(s)+2 \mathrm{NaOH}(a q)\] for the removal of ozone by sodium iodide, has already been balanced. Balanced equations have equal numbers of each type of atom on both sides of the arrow. This balance is essential for determining how much of each reactant is needed to produce a certain amount of product—a fundamental concept in stoichiometry called the stoichiometric ratio.

In our example, the stoichiometric ratio between ozone (O3) and sodium iodide (NaI) is 1:2. This means that for every one mole of ozone, two moles of sodium iodide are required to completely react. Without such balanced equations, we would be unable to proceed with accurate calculations and predictions in our chemical reactions.

Mole Concept

The mole concept is a gateway to understanding chemical quantities in reactions. Essentially, a mole represents Avogadro's number (\(6.022 \times 10^{23}\) entities) of any substance. This concept allows chemists to count atoms, molecules, or ions by weighing, a practical approach unlike counting each particle individually.

For the EPA method for determining ozone concentration, this concept allows us to convert between mass, moles, and the number of particles. When we know that we have, for instance, \(5.95 \times 10^{-6} mol\) of ozone, we understand that this represents approximately \(3.58 \times 10^{18}\) ozone molecules, as each mole contains Avogadro's number of particles. Moles offer a bridge between the macroscopic world we can measure and the microscopic world of atoms and molecules.

Molar Mass

To move from moles to grams or vice versa, we need the concept of molar mass—the mass of one mole of a substance. The molar mass is numerically equal to the element's atomic weight (from the periodic table) in grams per mole. For compounds, the molar mass is the sum of the atomic weights of all atoms in the molecule.

For example, to calculate the molar mass of sodium iodide (NaI), we add the atomic weights of sodium (22.99 g/mol) and iodine (126.90 g/mol), yielding a molar mass of 149.89 g/mol. In stoichiometric calculations, this allows us to convert moles of NaI into grams, which is a more useful measurement for practical applications like weighing out the substance in a laboratory. Thus, understanding molar mass is a cornerstone of performing accurate stoichiometric calculations.