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.
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 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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