Question 5.13: Aqueous lithium hydroxide solution is used to purify air in ...
Aqueous lithium hydroxide solution is used to purify air in spacecraft and submarines because it absorbs carbon dioxide, which is an end product of metabolism, according to the equation
2LiOH(aq) + CO_{2}(g) → Li_{2}CO_{3}(aq) + H_{2}O(l)The pressure of carbon dioxide inside the cabin of a submarine having a volume of 2.4 \times 10^{5} L is 7.9 \times 10^{-3} atm at 312 K. A solution of lithium hydroxide (LiOH) of negligible volume is introduced into the cabin. Eventually the pressure of CO_{2} falls to 1.2 \times 10^{-4} atm. How many grams of lithium carbonate are formed by this process?
Strategy How do we calculate the number of moles of CO_{2} reacted from the drop in CO_{2} pressure? From the ideal gas equation, we write
n=P\times (\frac{V}{RT} )At constant T and V, the change in pressure of CO_{2}, \Delta P, corresponds to the change in the number of moles of CO_{2}, \Delta n. Thus,
\Delta n=\Delta n\times (\frac{V}{RT} )What is the conversion factor between CO_{2} and Li_{2}CO_{3}?
The drop in CO_{2} pressure is (7.9 \times 10^{-3} atm) − (1.2 \times 10^{-4} atm) or 7.8 \times 10^{-3} atm. Therefore, the number of moles of CO_{2} reacted is given by
\Delta n=7.9 \times 10^{-3} atm \times \frac{2.4\times 10^{5} L}{(0.0821 L . atm/K . mol)(312 K)}=73 mol
From the chemical equation, we see that 1 mol CO_{2} \bumpeq 1 mol Li_{2}CO_{3} , so the amount of Li_{2}CO_{3} formed is also 73 moles. Then, with the molar mass of Li_{2}CO_{3} (73.89 g), we calculate its mass.
mass of Li_{2}CO_{3} formed =73 \cancel{mol Li_{2}CO_{3}}\times \frac{73.89 g Li_{2}CO_{3}}{1 \cancel{mol Li_{2}CO_{3}}}
=5.4 \times 10^{3} g Li_{2}CO_{3}