The astronauts of the Apollo 13 mission to the moon were almost lost in space due to the explosion of a valve on their primary oxygen generator. The lives of the astronauts were threatened by increasing concentrations of carbon dioxide in the spacecraft. Gaseous carbon dioxide(CO₂) be absorbed from the air in a spacecraft with open canisters that contain solid lithium hydroxide according to the following equation:
2 LiOH (s) + CO₂(g) $\rightarrow$ Li₂CO₃(s) + H₂O (l)
If a canister absorbed 225.0 grams of carbon dioxide from the air that resulted in the production of 66.30 grams of H₂O, what was the percent yield for H₂O?

Question

The astronauts of the Apollo 13 mission to the moon were almost lost in space due to the explosion of a valve on their primary oxygen generator. The lives of the astronauts were threatened by increasing concentrations of carbon dioxide in the spacecraft. Gaseous carbon dioxide(CO₂) be absorbed from the air in a spacecraft with open canisters that contain solid lithium hydroxide according to the following equation:
2 LiOH (s) + CO₂(g)

\rightarrow

Li₂CO₃(s) + H₂O (l)
If a canister absorbed 225.0 grams of carbon dioxide from the air that resulted in the production of 66.30 grams of H₂O, what was the percent yield for H₂O?

Quizplus · Accepted Answer

The percent yield is calculated using the formula: $\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$. First, we need to calculate the theoretical yield of $H_2O$ based on the given reaction and the amount of $CO_2$ absorbed. The molar mass of $CO_2$ is approximately 44.01 g/mol, and the molar mass of $H_2O$ is approximately 18.015 g/mol. From the balanced chemical equation, 1 mole of $CO_2$ produces 1 mole of $H_2O$. Therefore, 225.0 grams of $CO_2$ is equivalent to $\frac{225.0}{44.01}$ moles, which theoretically should produce the same number of moles of $H_2O$. The theoretical yield of $H_2O$ in grams is this mole amount multiplied by the molar mass of $H_2O$. The actual yield of $H_2O$ is given as 66.30 grams. Calculating the percent yield using these values gives a result close to 71.96%, indicating that the reaction's efficiency in producing $H_2O$ was about 71.96% of the theoretical maximum.

The Astronauts of the Apollo 13 Mission to the Moon →\rightarrow→

The Astronauts of the Apollo 13 Mission to the Moon $\rightarrow$