Question

A method used by the 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: O3(g)+2NaI(aq)+H2O(l)→O2(g)+I2(s)+2NaOH(aq)

How many grams of sodium iodide are needed to remove 1.1 mg of O3?
Express your answer using two significant figures.

Answer 1

Approximately 6.9 mg of sodium iodide are needed to remove 1.1 mg of ozone, calculated using stoichiometry and molar mass conversions.

Step-by-step explanation:

To determine how many grams of sodium iodide (NaI) are needed to remove 1.1 mg of ozone (O3), we can first look at the balanced chemical equation:

O3(g) + 2NaI(aq) + H2O(l) → O2(g) + I2(s) + 2NaOH(aq)

From the equation, we see that one molecule of ozone reacts with two molecules of sodium iodide. To find out how much sodium iodide is required, we'll need to perform a series of unit conversions and use the molar masses of ozone (O3) and sodium iodide (NaI).

First, we convert the mass of ozone from milligrams to grams:

1.1 mg O3 = 0.0011 g O3

Next, we need to find the moles of ozone using its molar mass (O3 has a molar mass of approximately 48 g/mol).

Moles of O3 = 0.0011 g O3 / 48 g/mol = 2.29 x 10-5 mol

According to the stoichiometry of the given equation, 1 mole of O3 requires 2 moles of NaI. So, the moles of NaI needed are:

Moles of NaI = 2 x 2.29 x 10-5 mol = 4.58 x 10-5 mol

We now convert moles of NaI to grams using its molar mass (NaI has a molar mass of approximately 150 g/mol):

Mass of NaI = 4.58 x 10-5 mol x 150 g/mol = 6.87 x 10-3 g

To finish, we convert this mass into milligrams and round to two significant figures:

Mass of NaI = 6.87 x 10-3 g x 1000 mg/g ≈ 6.9 mg

6.9 mg of sodium iodide are needed to remove 1.1 mg of ozone.