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) + CO2(g) → Li2CO3(aq) + H2O(l)The pressure of carbon dioxide inside the cabin of a submarine having a

Question

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

[latex]2LiOH(aq) + CO_{2}(g) → Li_{2}CO_{3}(aq) + H_{2}O(l)[/latex]

The pressure of carbon dioxide inside the cabin of a submarine having a volume of [latex]2.4 imes 10^{5} L is 7.9 imes 10^{-3}[/latex] atm at 312 K. A solution of lithium hydroxide (LiOH) of negligible volume is introduced into the cabin. Eventually the pressure of [latex]CO_{2}[/latex] falls to[latex] 1.2 imes 10^{-4}[/latex] atm. How many grams of lithium carbonate are formed by this process?

Strategy How do we calculate the number of moles of [latex]CO_{2}[/latex] reacted from the drop in [latex]CO_{2}[/latex] pressure? From the ideal gas equation, we write

[latex]n=P imes (\frac{V}{RT} )[/latex]

At constant T and V, the change in pressure of [latex]CO_{2}, \Delta P[/latex], corresponds to the change in the number of moles of [latex]CO_{2}, \Delta n[/latex]. Thus,

[latex]\Delta n=\Delta n imes (\frac{V}{RT} )[/latex]

What is the conversion factor between [latex]CO_{2}[/latex] and [latex]Li_{2}CO_{3}[/latex]?

Accepted Answer

The drop in [latex]CO_{2}[/latex] pressure is ([latex]7.9 imes 10^{-3}[/latex] atm) − ([latex]1.2 imes 10^{-4}[/latex] atm) or [latex]7.8 imes 10^{-3}[/latex] atm. Therefore, the number of moles of [latex]CO_{2}[/latex] reacted is given by

[latex]\Delta n=7.9 imes 10^{-3} atm imes \frac{2.4 imes 10^{5} L}{(0.0821 L . atm/K . mol)(312 K)} [/latex]

=73 mol

From the chemical equation, we see that 1 mol [latex]CO_{2} \bumpeq 1 mol Li_{2}CO_{3}[/latex] , so the amount of [latex]Li_{2}CO_{3}[/latex] formed is also 73 moles. Then, with the molar mass of [latex] Li_{2}CO_{3}[/latex] (73.89 g), we calculate its mass.

mass of [latex]Li_{2}CO_{3} formed =73 \cancel{mol Li_{2}CO_{3}} imes \frac{73.89 g Li_{2}CO_{3}}{1 \cancel{mol Li_{2}CO_{3}}} [/latex]

[latex]=5.4 imes 10^{3} g Li_{2}CO_{3}[/latex]

Question 5.13: Aqueous lithium hydroxide solution is used to purify air in ...

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