Question

A method used by the U.S. Environmental Protection Agency \((\) EPA) for determining the concentration of ozone in airis to pass the air sample through a "bubbler" containing sodium iodide, which removes the ozone according to the following equation: $$ \begin{array}{c}{\mathrm{O}_{3}(g)+2 \operatorname{NaI}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow} \\ \quad \quad \quad \quad \quad \quad\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad{\mathrm{O}_{2}(g)+\mathrm{I}_{2}(s)+2 \mathrm{NaOH}(a q)}\end{array} $$ (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 \(\mathrm{O}_{3} ?\)

Short answer

(a) \(1.19 \times 10^{-5} \ \text{moles}\) of sodium iodide are needed to remove \(5.95 \times 10^{-6}\ \mathrm{mol}\) of ozone. (b) \(8.13 \times 10^{-3}\ \text{g}\) of sodium iodide are needed to remove 1.3 \(\mathrm{mg}\) of ozone.

Step-by-step solution

Part (a) - Find moles of sodium iodide

Using the balanced equation, we can see that 1 mole of ozone reacts with 2 moles of sodium iodide. So, we can set up the conversion factor: $$ \frac{2 \ \text{moles of NaI}}{1\ \text{mole of O}_3 } $$ Now multiply the given moles of ozone with the conversion factor to find the needed moles of sodium iodide: $$ (5.95 \times 10^{-6}\text{ mol of O}_3) \times \frac{2\ \text{moles of NaI}}{1\ \text{mole of O}_3 } $$

Part (a) - Calculate the moles of sodium iodide

Simply multiply the two values together: $$ (5.95 \times 10^{-6})\times 2 = 1.19 \times 10^{-5} \ \text{moles of NaI} $$ So, \(1.19 \times 10^{-5} \ \text{moles}\) of sodium iodide are needed to remove \(5.95 \times 10^{-6}\ \mathrm{mol}\) of ozone.

Part (b) - Convert mass of ozone to moles

First, let's convert the given mass of ozone (1.3 mg) to moles. To do this, we will use the molar mass of ozone (48 g/mol): $$ 1.3\ \mathrm{mg} \times \frac{1\ \mathrm{g}}{1000\ \mathrm{mg}} \times \frac{1\ \mathrm{mol}}{48\ \mathrm{g}} $$

Part (b) - Calculate moles of ozone

Perform the multiplication and division: $$ \frac{1.3}{1000 \times 48} = 2.71 \times 10^{-5}\ \text{mol} $$

Part (b) - Find moles of sodium iodide

Now we can use the same conversion factor from (a): $$ (2.71 \times 10^{-5}\ \text{mol of O}_3) \times \frac{2\ \text{moles of NaI}}{1\ \text{mole of O}_3 } $$

Part (b) - Calculate moles of sodium iodide

Multiply the values: $$ (2.71 \times 10^{-5})\times 2 = 5.42 \times 10^{-5} \ \text{moles of NaI} $$

Part (b) - Convert moles of sodium iodide to grams

Finally, we need to convert the moles of sodium iodide to grams. The molar mass of sodium iodide is approximately 150 g/mol: $$ (5.42 \times 10^{-5}\ \text{mol of NaI}) \times \frac{150\ \mathrm{g}}{1\ \mathrm{mol}} $$

Part (b) - Calculate grams of sodium iodide

Perform the multiplication: $$ (5.42 \times 10^{-5}) \times 150 = 8.13 \times 10^{-3} \ \text{g} $$ So, \(8.13 \times 10^{-3}\ \text{g}\) of sodium iodide are needed to remove 1.3 \(\mathrm{mg}\) of ozone.