The CO2 that builds up in the air of a submerged submarine can be removed by reacting it with sodium peroxide. 2 Na2O2 (s) + 2 CO2 (g) → 2 Na2CO3 (s) + O2 (g)
If a sailor exhales 150.0 mL of CO2 per minute at 20oC and 1.02 atm, how much sodium peroxide is needed per sailor in a 24 hr period?​

Answer: The sodium peroxide needed per sailor in a 24 hr period is 674 grams.

Explanation:

According to ideal gas equation:

[tex]PV=nRT[/tex]

P = pressure of gas = 1.02 atm

V = Volume of gas = 150.0 ml = 0.15 L

n = number of moles  of carbon dioxide = ?

R = gas constant =[tex]0.0821Latm/Kmol[/tex]

T =temperature =[tex]20^0C=(20+273)K=293K[/tex]

[tex]n=\frac{PV}{RT}[/tex]

[tex]n=\frac{1.02atm\times 0.15L}{0.0821 L atm/K mol\times 293K}=0.006moles[/tex]

[tex]2Na_2O_2(s)+2CO_2(g)\rightarrow 2Na_2CO_3(s)+O_2(g)[/tex]

According to stoichiometry:

2 moles of carbon dioxide require = 2 moles of sodium peroxide

Thus 0.006 moles carbon dioxide require= [tex]\frac{2}{2}\times 0.006=0.006[/tex] moles of  sodium peroxide

Mass of  sodium peroxide needed for 24 hour=[tex]moles\times {\text {Molar mass}}=0.006mol\times 78g/mol\times 24\times 60=674g[/tex]

Thus sodium peroxide needed per sailor in a 24 hr period is 674 grams.

The amount (in grams) of sodium peroxide needed per sailor in a 24 hr period is 673.92 g

Using the ideal gas equation, we can obtain the mole of CO₂ as follow:

n = PV / RT

n = (1.02 × 0.15) / (0.0821 × 293)

n = 0.006 mole

2Na₂O₂ + 2CO₂ → 2Na₂CO₃ + O₂

From the balanced equation above,

2 moles of CO₂ required 2 moles of Na₂O₂.

Therefore,

0.006 mole of CO₂ will also require 0.006 mole of Na₂O₂.

Mass = mole × molar mass

Mass of Na₂O₂ = 0.006 × 78

Mass of Na₂O₂ = 0.468 g per minute

Mass of Na₂O₂ required for 24 h = 0.468 × 24 × 60 = 673.92 g

Learn more about stoichiometry:

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