An environmental engineer analyzes a sample of air contaminated with sulfur dioxide. To a 500.-mL sample at 700 . torr and \(38^{\circ} \mathrm{C},\) she adds \(20.00 \mathrm{~mL}\) of \(0.01017 M\) aqueous iodine, which reacts as follows: \(\mathrm{SO}_{2}(g)+\mathrm{I}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) $$\mathrm{HSO}_{4}^{-}(a q)+\mathrm{I}^{-}(a q)+\mathrm{H}^{+}(a q) \quad[\text { unbalanced }]$$ Excess \(\mathrm{I}_{2}\) reacts with \(11.37 \mathrm{~mL}\) of \(0.0105 \mathrm{M}\) sodium thiosulfate: \(\mathrm{I}_{2}(a q)+\mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q) \longrightarrow \mathrm{I}^{-}(a q)+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}(a q) \quad[\) unbalanced \(]\) What is the volume \(\%\) of \(\mathrm{SO}_{2}\) in the air sample?

In this exercise, the ideal gas law serves as a crucial tool. The ideal gas law relates pressure, volume, temperature, and the amount of gas through the equation: \(PV = nRT\). Here, P stands for pressure, V for volume, n for moles, R for the gas constant, and T for temperature.
By manipulating this formula, you can solve for any one missing variable if the other variables are known.
For example, in Step 4 of our problem, we rearrange the ideal gas law to solve for volume (V) of sulfur dioxide (SO\textsubscript{2}): \(V = \frac{nRT}{P}\).\
Make sure to convert your measurements to the correct units: pressure to atmospheres (atm), volume to liters (L), temperature to Kelvin (K), and use the gas constant R = 0.0821 atm•L/mol•K.

Stoichiometry involves the calculation of reactants and products in chemical reactions. It is based on the law of conservation of mass. In this problem, two reactions are involved: \(\mathrm{SO_{2} (g) + I_{2} (aq) + H_{2}O (l) \rightarrow HSO_{4}^{-} (aq) + I^{-} (aq) + H^{+}(aq)}\) and \(\mathrm{I_{2}(aq) + S_{2}O_{3}^{2-}(aq) \rightarrow I^{-}(aq) + S_{4}O_{6}^{2-}(aq)}\). Understanding the stoichiometry of these reactions allows us to determine the amount of sulfur dioxide reacting with iodine.
For instance, the 1:1 stoichiometric ratio of \(I_{2}\) reacting with \(S_{2}O_{3}^{2-}\) helps in calculating the excess \(I_2\). By knowing the moles of \(Na_{2}S_{2}O_{3}\), and using this ratio, we determine the moles of leftover \(I_{2}\). Such stoichiometric relationships are vital for solving various parts of the problem.

Moles calculation is essential for understanding the quantities involved in chemical reactions.
In the context of this exercise, it all started with calculating moles of \(\mathrm{I_{2}}\) added: \(n = C \times V\), where C is concentration and V is volume. The determined moles of added \(I_{2}\) sets the frame for subsequent calculations.
The next step involved calculating the moles of sodium thiosulfate (\(\mathrm{Na_{2}S_{2}O_{3}}\)) that have reacted with the excess iodine: again using \(n = C \times V\). These values were essential to determine the moles of \( I_2 \) actually reacting with sulfur dioxide by simple subtraction.
Without understanding the basic moles calculation, moving forward with the rest of the problem would be very challenging.

Sulfur dioxide (SO\(\mathrm{_2}\)) analysis involves understanding its interaction in the given chemical reactions.
Initially, the engineer added iodine to the air sample to react with all the sulfur dioxide present. \( \mathrm{SO_{2} (g) + I_{2} (aq) + H_{2}O (l) \rightarrow HSO_{4}^{-} (aq) + I^{-} (aq) + H^{+}(aq)} \). Later, the unreacted \(\mathrm{I_{2}}\) was treated with sodium thiosulfate (\(\mathrm{Na_{2}S_{2}O_{3}}\)) allowing the calculation of the iodine initially reacted.
Once molecular quantities were determined, the concentration of sulfur dioxide in the air sample could be calculated and expressed as a volume percentage.
This type of analysis helps the environmental engineer assess air quality and understand pollution levels efficiently.