fter lithium hydroxide is produced aboard the space shuttle by reaction of Li2O

Question

fter lithium hydroxide is produced aboard the space shuttle by reaction of Li2O with H2O, it is used to remove exhaled carbon dioxide from the air supply. Initially 327.1 g of LiOH were present and 542.1 g of LiHCO3 have been produced. LiOH(s)+CO2(g)LiHCO3(s) Part A Can the reaction remove any additional CO2 from the air? Additional CO2 can be removed. Additional CO2 cannot be removed. SubmitMy AnswersGive Up Correct Part B If so, how much? Express your answer to four significant figures and include the appropriate units.

Explanation / Answer

Calculate moles of LiOH first:
moles = 327.1 g / 23.94 = 13.6633 moles

moles of CO2 = 13.6633 moles (by stechiometry)
mass of CO2 = 13.6633*44 = 601.1852 g

These moles would be the same moles produces of LiHCO3, so the mass is:
mass of product = 13.6633 * (6.94+1+12+48) = 928.0604 g produced

The mass produced was 542,1 g with the 327.1 g of LiOH, so the mass of CO2 absorbed was:
mCO2 = 542.1 - 327.1 = 215 g

Which means that according to the innitial data, and innitial mass, it cannot removed more CO2.

Hope this helps

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