Question

An air-purification method for enclosed spaces involves the use of "scrubbers" containing aqueous lithium hydroxide, which reacts with carbon dioxide to produce lithium carbonate and water: 2LiOH(aq) + CO2(g) → Li2CO3(s) + H2O(l) Consider the air supply in a submarine with a total volume of 2.4 × 105 L. The pressure is 0.9970 atm, and the temperature is 25°C. By how much would the pressure in the submarine drop if 3.15 kg of LiOH were completely consumed by reaction with CO2?

Answer 1

2LiOH(aq) + CO2(g) ---> Li2CO3(s) + H2O(l)

from equation , 2 mol LiOH = 1 mol CO2

no of mole of air in submarine(n) = PV/RT

                   = 0.997*2.4*10^5/(0.0821*298)

                   = 9780.2 mole

no of mole of LiOH allowed = w/mwt = 3.15*10^3/23.95 = 131.52 mole

NO of mole of CO2 reacts = 131.52/2 = 65.76 mole

so that , no of mole of Air in submarine reducedto = 9780.2-65.76 = 9714.44 mole

final pressure(P) = nRT/V

              P = 9714.44*0.0821*298/(2.4*10^5)

                = 0.991 atm

change in pressure(DP) = 0.997 -0.991 = 0.006 atm