Hello, Angeleen,
__2__ LiOH(s) + __1__ CO2 (g)→ _1___Li2CO3 (s) + __1_ H2O (l)
1.The filter aboard the spaceship contains 265 moles of lithium hydroxide. How many moles of carbon dioxide will this amount remove from the spaceship’s cabin?
The balanced equation says 2 moles of LiOH will react with only 1 mole of CO2. Therefore, 265 moles of LiOH will consume (1/2)(265 moles) = 132.5 moles of CO2
2.The astronauts have two (2) days left before they land on Earth. There are three (3) astronauts, and each astronaut emits approximately 15 moles of carbon dioxide each day. How much carbon dioxide will they expel during the two (2) day flight back to Earth?
(3 astronauts)(15 moles CO2/day/astronaut)*(2 days) = 90 moles CO2
3.Is there enough lithium hydroxide in the filter to cleanse the cabin air of the carbon dioxide, or are the astronauts doomed? Support your answer with your calculations and answers from above.
As per the answer to 1), there is enough LiOH to cleanse the air for the 2 day return to Earth.
4.If the astronauts survive, they are not out of the woods yet! The spacecraft is sensitive to imbalances in weight when it re-enters the Earth’s atmosphere, therefore they need to be aware of the waste produced when they use the lithium hydroxide filter. The water from this reaction can be released prior to entry, however the solid lithium carbonate will need to be stored. The spacecraft can only account for an additional 7.5 kg before the balance will be altered causing landing problems. Can the astronauts re-enter the Earth’s atmosphere safely or is too much solid waste produced when using the lithium hydroxide filter?
[ 1 mole of Li2CO3 = 75 grams Li2CO3 ]
This is an interesting question. I'll do the math to determine how much Li2CO3 mass is formed for the remaining trip of 2 days. Then I'll provide the more interesting answer.
90 moles of CO2 are captured in the reaction to form Li2CO3. The balanced equation says we'll get 1 mole Li2CO3 for every 1 mole of CO2 consumed. That means 90 moles of Li2CO3 will form over the two days.
(90 moles LiCO3)*(75g/mole Li2CO3) = 6750 g LiCO3 formed. That's 6.75 kgof Li2CO3, under the 7.5kg limit, assuming the H2O can be expelled.
BUT, the more interesting analysis, made by an undoubtedly shrewd observer, would have been to answer this question with no calculations. This observer, we'll call him Spock, declares that there will be no problem with additional mass for the remainder of the journey. The CO2 captured by the filter is already aboard the spacecraft, tied up among the many molecules in the human body. The tubby astronaut in the corner, eating Twinkies, has a lot of carbohydrates just waiting to be decomposed in a biochemical pathway to release energy and byproducts of CO2 and H2O. So the CO2 is already aboard the spacecraft - there is no change in mass, just a change in form. In fact, he muses, the H2O molecules are also on board. Ejecting them into the Martian wilderness would, in fact, reduce the mass of the spacecraft. A dangerous proposition since the flight trajectory depends on the current mass. Alas, we read about Spock's dire warning as the spaceship is headed for the asteroid belt after going off course soon after the ship's android purged the wastewater tanks as per Houston's preprogrammed instructions, bringing the craft below critical mass for reentry.
I hope this humors helps.
Bob
