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 1.6 × 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 2.35 kg of LiOH were completely consumed by reaction with CO2?

atm

Answer 1

Ans. 2 LiOH _(aq) + CO_2(aq) ------------------> Li₂CO_3(s) + H₂O_(l)

Stoichiometry: 2 moles LiOH reacts with 1 mol CO₂.

Given,

            Mass of LioH consumed = 2.35 Kg = 2350 g

Moles of LioH consumed = Mass of LiOH consumed / molar mass of LiOH

                                    = 2350 g / 23.947 g mol^-1

                                    = 98.13 moles

Now, moles of CO₂ = half the moles of LiOH

                        = ½ x 98.13 moles = 49.066 moles

Part 2: Calculate pressure contributed by 49.066 moles CO₂

Using ideal gas equation Ideal gas Law: pV = nRT

            Where, p = pressure in atm = ?

            V = volume in L             = 1.6 x 10⁵ L

            n = number of moles = 49.066

            R = universal gas constant= 0.082057338 atm L mol^-1K^-1

            T = absolute temperature           = 298.15 K                    (0⁰C = 273.15 K)

Putting the values in above equation

P = nRT / V = [ 49.066 moles x (0.082057338 atm L mol^-1K^-1) x 298.15 K ] / 1.6 x 10⁵ L

Or, P = 0.0074989 atm

That is, the pressure exerted by 49.066 moles (absorbed by LiOH) is equal to 0.0074989 atm.

Thus, absorption of 49.066 moles CO₂ drops the pressure by 0.0074989 atm. units.