Question

The Apollo 13 mission was determined to be a “successful failure”. Although the astronauts did not land on the Moon, they were successfully brought home. A thorough investigation of the explosion, based on records of manufacturing processes and maintenance logs, tracked the failure of the oxygen tank to multiple faults. None of the problems, had they occurred alone, would have been critical, but together they led to a near disaster for the crew of Apollo 13. How much carbon dioxide would have been removed from the breathing environment if the scrubber reaction produced 50.0 kilograms of lithium carbonate? Use dimensional analysis.

Answer 1

By using dimensional analysis with the chemical reaction of carbon dioxide and lithium hydroxide, we can determine that approximately 29.7 kg of CO2 would have been removed if the scrubber reaction produced 50.0 kg of lithium carbonate.

Step-by-step explanation:

To calculate how much carbon dioxide (CO2) would be removed from the Apollo 13 crew's breathing environment if 50.0 kilograms of lithium carbonate (Li2CO3) were produced, we can use dimensional analysis with the reaction equation between carbon dioxide and lithium hydroxide (LiOH), which was used in the scrubbers on Apollo 13:

2 LiOH + CO2 → Li2CO3 + H2O

The molar mass of Li2CO3 is approximately 73.89 g/mol, and the molar mass of CO2 is about 44.01 g/mol. Using dimensional analysis:

50,000 g Li2CO3 × (1 mol Li2CO3 / 73.89 g Li2CO3) × (1 mol CO2 / 1 mol Li2CO3) × (44.01 g CO2 / 1 mol CO2) = 29,713.55 g CO2 or approximately 29.7 kg of CO2 removed.

Answer 2

29,769.95g of carbon dioxide can be removed.

1st) It is necessary to write the balanced chemical reaction:

`2\text{LiOH + CO}_2\text{ }\rightarrow Li_2CO_3+H_2O`

According to the balanced equation, 1 mole of carbon dioxide (CO2) can be removed by producing 1 mole of lithium carbonate (Li2CO3).

Using the molar mass of carbon dioxide and lithium carbonate we can convert moles into grams:

- CO2 molar mass: 44 g/mol

- Li2CO3 molar mass: 73.9 g/mol

With the molar mass we can see that 44g of carbon dioxide (1 mole) can be removed by producing 73.9 g of lithium carbonate.

2nd) Now we can calculate the amount of carbon dioxide that can be removed from the breathing environment if the scrubber reaction produces 50.0 Kg of Li2CO3, using a mathematical rule of three:

`\begin{gathered} 0.0739kgLi_2CO_3-44gCO_2 \\ 50kgLi_2CO_3-x=(50kgLi_2CO_3\cdot44gCO_2)/(0.0739kgLi_2CO_3) \\ x=29,769.95gCO_2 \end{gathered}`

Here, it is important to convert the grams of Li2CO3 to kg before doing the calculus.

So, 29,769.95g (29.8kg) of carbon dioxide can be removed.