The space shuttle environmental control system handles excess \(\mathrm{CO}_{2}\) (which the astronauts breathe out; it is \(4.0 \%\) by mass of exhaled air) by reacting it with lithium hydroxide, LiOH, pellets to form lithium carbonate, \(\mathrm{Li}_{2} \mathrm{CO}_{3}\), and water. If there are 7 astronauts on board the shuttle, and each exhales \(20 . \mathrm{L}\) of air per minute, how long could clean air be generated if there were \(25,000 \mathrm{~g}\) of LiOH pellets available for each shuttle mission? Assume the density of air is \(0.0010 \mathrm{~g} / \mathrm{mL}\).

We know each astronaut exhales 20 L of air per minute and \(\mathrm{CO}_{2}\) is 4.0% by mass of the exhaled air. We are also given the density of air as 0.0010 g/mL. First, we need to find the mass of the exhaled air and then calculate the mass of \(\mathrm{CO}_{2}\) in that air. Mass of exhaled air per minute by one astronaut = Volume × Density = 20 L × 1000 mL/L × 0.0010 g/mL = 20 g Now, let's calculate the mass of \(\mathrm{CO}_{2}\) exhaled by one astronaut per minute: Mass of \(\mathrm{CO}_{2}\) = (Mass of exhaled air × Percentage of \(\mathrm{CO}_{2}\)) / 100 = (20 g × 4.0) / 100 = 0.8 g

Since there are 7 astronauts on board, we need to calculate the total mass of \(\mathrm{CO}_{2}\) exhaled by all of them per minute: Total mass of \(\mathrm{CO}_{2}\) per minute = Mass of \(\mathrm{CO}_{2}\) per astronaut × Number of astronauts = 0.8 g × 7 = 5.6 g

Next, we need to calculate the number of moles of \(\mathrm{CO}_{2}\) exhaled per minute and the moles of LiOH provided: Moles of \(\mathrm{CO}_{2}\) per minute \(= \frac{\text{mass of}\ \mathrm{CO}_{2}\ \text{per minute}}{\text{molar mass of}\ \mathrm{CO}_{2}}\) Moles of \(\mathrm{CO}_{2}\) per minute \(= \frac{5.6\ g}{(12.01\ g/mol\ \text{for C}) + (2 \times 16.00\ g/mol\ \text{for O})} = \frac{5.6\ g}{44.01\ g/mol} \approx 0.127\) mol Moles of LiOH \(= \frac{\text{mass of LiOH}}{\text{molar mass of LiOH}}\) Moles of LiOH \(= \frac{25000 g}{(6.94\ g/mol\ \text{for Li}) + (15.99\ g/mol\ \text{for O}) + (1.008\ g/mol\ \text{for H})} = \frac{25000\ g}{23.948\ g/mol} \approx 1044\) mol

Using the balanced chemical equation, \(2\text{LiOH} + \mathrm{CO}_{2} \rightarrow \mathrm{Li}_{2}\mathrm{CO}_{3} + \text{H}_{2}\text{O}\), we can deduce that 2 moles of LiOH reacts with 1 mole of \(\mathrm{CO}_{2}\). Therefore, let's find how many moles of \(\mathrm{CO}_{2}\) can react with the available LiOH: Moles of \(\mathrm{CO}_{2}\) reacting with the available LiOH = \(\frac{1}{2} \times\) Moles of LiOH = \(\frac{1}{2} \times 1044 \text{ moles} = 522\) mol Now, let's calculate how many minutes this amount of LiOH can generate clean air, considering the moles of \(\mathrm{CO}_{2}\) exhaled per minute: Minutes to generate clean air \(= \frac{\text{total moles of}\ \mathrm{CO}_{2}\ \text{reacting with LiOH}}{\text{moles of}\ \mathrm{CO}_{2}\ \text{exhaled per minute}}\) Minutes to generate clean air \(= \frac{522 \text{ mol}}{0.127 \text{ mol/min}} \approx 4110\) minutes Hence, the 25,000 g of LiOH pellets can generate clean air for approximately 4110 minutes if there are 7 astronauts on board.