Question

The space shuttle environmental control system handled excess${\mathbf{\text{CO}}}_{\mathbf{2}}$ (which the astronauts breathe out; it is $\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\%}$ by mass of exhaled air) by reacting it with lithium hydroxiderole="math" localid="1648565378166" $\left(\text{LiOH}\right)$pelletsto form lithium carbonate $\left({\text{Li}}_2{\text{CO}}_3\right)$and water. If there were seven astronauts on board the shuttle, and each exhales 20. L of air per minute, how long could clean air be generated if there were $\mathbf{25}\mathbf{,}\mathbf{000}\mathbf{\text{g}}$of $\left(\text{LiOH}\right)$ pellets available for each shuttle mission? Assume the density of air is$\mathbf{0}\mathbf{.}\mathbf{0010}\mathbf{}\raisebox{1ex}{$\mathbf{\text{g}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{m}\mathbf{L}$}\right.$ .

Short answer

Clean air will be generated$4108$ minutes once, which is 2.85 days, or 2 days, 20 hours, 18 minutes, and 21 seconds.

Step-by-step solution

Definition

A balanced chemical equation is one in which the number of atoms present on the reactant side of the equation equals the number of atoms present on the product side of the equation.

Balancing the given equation 

The balanced equation is as follows

${\text{CO}}_2\text{+2Li}\left(\text{OH}\right)\text{→Li}\left({\text{CO}}_3\right){\text{+H}}_2\text{O}$

Formation of Stoich table

Each passenger uses$20$liters of air per minute, so $20\times 7$ becomes $140$litersper minute, and then you subtract $4\%$ to get $5.6$litersper minute ${\text{CO}}_2$. Then you create a stoich table as follows:

role="math" localid="1648564286284" $=\text{5.6L}\raisebox{1ex}{${\text{CO}}_2$}\!\left/ \!\raisebox{-1ex}{$\text{min}$}\right.\text{×1000}\raisebox{1ex}{$\text{ml}$}\!\left/ \!\raisebox{-1ex}{$\text{1L}$}\right.\text{×0.0010g}\raisebox{1ex}{${\text{CO}}_2$}\!\left/ \!\raisebox{-1ex}{$\text{1ml}$}\right.$

$=1\text{mol}\raisebox{1ex}{${\text{CO}}_2$}\!\left/ \!\raisebox{-1ex}{$44.01\text{g}$}\right.\text{×2mol}\raisebox{1ex}{$\text{LiOH}$}\!\left/ \!\raisebox{-1ex}{$1{\text{molCO}}_2$}\right.$$\times \frac{1}{25000\text{g}}$

$=2.4{\text{e}}^{-4}{\text{min}}^{-1}$

Inverse the above obtained value and will get minutes.

Therefore, it takes$4108$ minutes for clean air to be generated. This equates to approximately 2.85 days, or 2 days, 20 hours, 18 minutes, and 21 seconds.