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: $\mathrm{O}_{3}(g)+2 \mathrm{NaI}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow$ $$ \mathrm{O}_{2}(g)+\mathrm{I}_{2}(s)+2 \mathrm{NaOH}(a q) $$ (a) How many moles of sodium iodide are needed to remove $5.95 \times 10^{-6} \mathrm{~mol}\( of \)\mathrm{O}_{3} ?(\mathbf{b})$ How many grams of sodium iodide are needed to remove \(1.3 \mathrm{mg}\) of \(\mathrm{O}_{3}\) ?

The balanced chemical equation is: \[ O_3(g) + 2 NaI (aq) + H_2O(l) \rightarrow O_2(g) + I_2(s) + 2 NaOH(aq) \] From the equation, we can see that: 1 mol of O₃ reacts with 2 mol of NaI. Now, we will determine the number of moles of sodium iodide needed to remove 5.95 × 10⁻⁶ mol of O₃ using mole ratio calculations. Let x be the moles of NaI: \( \frac{5.95 \times 10^{-6} mol~ O_3}{1 mol~ O_3} = \frac{x~ moles~ NaI}{2 mol~ NaI} \) Now, solve for x: \( x = 2 \times (5.95 \times 10^{-6} mol~ O_3) = 11.9 \times 10^{-6} mol~ NaI \) So, 11.9 × 10⁻⁶ mol of sodium iodide are needed to remove 5.95 × 10⁻⁶ mol of ozone.

First, we need to convert the mass of ozone to moles, knowing that the molar mass of O₃ is 48 g/mol. \( \text{moles of }O_3 = \frac{1.3 mg}{1000} \times \frac{1 mol}{48 g} = 2.708 \times 10^{-5} mol \) Now, we can determine the number of moles of sodium iodide needed to remove 2.708 × 10⁻⁵ mol of O₃ using mole ratio calculations: \( \frac{2.708 \times 10^{-5} mol~ O_3}{1 mol~ O_3} = \frac{x~ moles~ NaI}{2 mol~ NaI} \) Now, solve for x: \( x = 2 \times (2.708 \times 10^{-5} mol~ O_3) = 5.416 \times 10^{-5} mol~ NaI \) Now convert the moles of sodium iodide to mass by multiplying by the molar mass of NaI (149.89 g/mol): \( mass~ NaI = 5.416 \times 10^{-5} mol~ NaI \times \frac{149.89 g~ NaI}{1 mol~ NaI} = 0.00813 g \) So, 0.00813 g of sodium iodide are needed to remove 1.3 mg of ozone.