Chapter 5
Q. 5.13
Q. 5.13
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}?
Step-by-Step
Verified Solution
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}