We are asked to estimate how much KO_{2} would be required for the apparatus.

4KO_{2}(s) + 2CO_{2}(g) → 2K_{2}CO_{3}(s) + 3O_{2}(g)

Step 1. Calculate the volume of oxygen to be replenished

$\mathbf{0}\mathbf{.}\mathbf{04}\mathbf{\times}\mathbf{0}\mathbf{.}\mathbf{20}\mathbf{\times}\frac{\mathbf{8}\mathbf{}\mathbf{L}\mathbf{}\mathbf{air}}{\mathbf{1}\mathbf{}\overline{)\mathbf{min}}}\mathbf{\times}\mathbf{13}\mathbf{}\overline{)\mathbf{min}}$

**= 0.832 L oxygen**

Step 2. Calculate moles O_{2}

$\overline{){\mathbf{PV}}{\mathbf{=}}{\mathbf{nRT}}}\phantom{\rule{0ex}{0ex}}\mathbf{n}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{PV}}{\mathbf{RT}}\phantom{\rule{0ex}{0ex}}\mathbf{n}\mathbf{}\mathbf{=}\mathbf{}\frac{(1\overline{)\mathrm{atm}})(0.832\overline{)L})}{(0.08206{\displaystyle \frac{\overline{)L}-\overline{)\mathrm{atm}}}{\mathrm{mol}-\overline{)K}}})(298.15\overline{)K})}$

**n = 0.034 mol O _{2}**

A particular kind of emergency breathing apparatus-often placed in mines, caves, or other places where oxygen might become depleted or where the air might become poisoned-works via the following chemical reaction:

4KO_{2}(s) + 2CO_{2}(g) → 2K_{2}CO_{3}(s) + 3O_{2}(g)

Notice that the reaction produces O_{2}, which can be breathed, and absorbs CO_{2}, a product of respiration. Suppose you work for a company interested in producing a self-rescue breathing apparatus (based on the above reaction) which would allow the user to survive for 13 minutes in an emergency situation.

Estimate how much KO_{2} would be required for the apparatus. (Find any necessary additional information. Assume that normal air is 20% oxygen and human breathing rate is about 8 Lair/min. Assume that 4% of the air would need to be replenished with oxygen.)

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