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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 ofc co2 in the space craft. Gaseous co2 can be absorbed from the air in a spacecraft with open canisters that contain solid lithium according to the following eqn
2 LIOH +CO2------>LI2 CO3 + H2O
If a canister absorbed 100.0grams of CO2 from the air that resulted in the production of 148.5grams of LI2 CO3, what was the percent yield for LI2 CO

Answer 1

as far as I got it figured 65%

Answer 2

Percent yield = Actual yield /theoredical yield

We are given the actual yield (148.5 g Li2CO3), so we must calculate the theoredical yield. For that, we need to do a gram to gram problem, discovering how many grams of Li2CO3 could be obtained from 100g of CO2.

100g CO2 x (1 mol CO2/44 g CO2) x (1 mol Li2CO3/1mol CO2) x (74g Li2CO3/1 mol Li2CO3) = 168.2 g Li2CO3 expected.

148.5/168.2 = 88.3% yield

Answer 3

Cutting out the background story, this is just a mole conversion problem.

Always, always, go through moles to convert one substance to another.

Each CO2 gives one L12CO3

100.0 g CO2 divided by formula mass CO2 gives moles CO2

Each mole CO2 gives one mole Li2CO3.

To get theoretical yield, multiply this number of moles by the formula mass of Li2CO3. This is your "theoretical yield".

Percent yield = actual yield/theoretical yield x 100%

Principles:

a) Ignore irrelevancies
b) Work in moles to go from one substance to another
c) Percentage = actual/theoretical

Good luck!

Answer 4

Good luck with that!!
i do a-level chemistry and i dont know!! rough guess although it is most likely wrong would be 48.5%

Answer 5

this is tough

Answer 6

i don't know