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Chapter 5

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}?

 

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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}