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## 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 $Li_{2}CO_{3}$? ## 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 $Li_{2}CO_{3} formed =73 \cancel{mol Li_{2}CO_{3}}\times \frac{73.89 g Li_{2}CO_{3}}{1 \cancel{mol Li_{2}CO_{3}}}$

$=5.4 \times 10^{3} g Li_{2}CO_{3}$