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 seven astronauts on board the shuttle, and each exhales 20. L of air per minute, how long could clean air be generated if there were 25,000 g of LiOH pellets available for each shuttle mission? Assume the density of air is 0.0010 g/mL.

Firstly, we need to determine the mass of CO₂ exhaled by seven astronauts each minute. To do this, we will use the given information that CO₂ comprises 4.0% by mass of the exhaled air and the density of air is 0.0010 g/mL. The volume of air exhaled per minute by one astronaut is 20 L, which is equivalent to 20,000 mL: Mass of air exhaled per minute by one astronaut = Volume x Density = 20,000 mL x 0.0010 g/mL = 20 g As there are seven astronauts, the total mass of air exhaled per minute is 7 x 20 g = 140 g. Since CO₂ is 4.0% by mass of the exhaled air: Mass of CO₂ exhaled per minute by seven astronauts = 0.040 x 140 g = 5.6 g

The balanced chemical equation representing the reaction between CO₂ and LiOH to form Li₂CO₃ and H₂O is: 2 LiOH + CO₂ → Li₂CO₃ + H₂O

From the balanced equation, we can see that 2 moles of LiOH react with 1 mole of CO₂. To find out how much LiOH is required to react with the CO₂ exhaled per minute, we need to convert the mass of CO₂ to moles and then use the stoichiometry to find the mass of LiOH: Molar mass of CO₂ = 12.01 g/mol (C) + 2 x 16.00 g/mol (O) = 44.01 g/mol Moles of CO₂ exhaled per minute = Mass / Molar Mass = 5.6 g / 44.01 g/mol ≈ 0.127 mol Now, use the balanced equation stoichiometry (2 moles of LiOH react with 1 mole of CO₂), we can find the moles of LiOH required: Moles of LiOH required = 2 x 0.127 mol ≈ 0.254 mol To find the mass of LiOH, we use the molar mass of LiOH (6.94 g/mol (Li) + 15.99 g/mol (O) + 1.01 g/mol (H) = 23.94 g/mol): Mass of LiOH required per minute = Moles x Molar Mass = 0.254 mol x 23.94 g/mol ≈ 6.08 g

We have been given that there are 25,000 g of LiOH pellets available for each shuttle mission. To find out how long it would last, divide the total available mass of LiOH by the mass required per minute: Duration (minutes) = Total mass of LiOH / Mass of LiOH required per minute = 25,000 g / 6.08 g/min ≈ 4,111 min To convert minutes to hours, divide the result by 60: Duration (hours) = 4,111 min / 60 min/h ≈ 68.5 h Thus, the available LiOH can generate clean air for approximately 68.5 hours for seven astronauts.