Answer:
725.15 L
Explanation:
The balanced chemical equation for the reaction between Na₂O₂ and CO₂ is the following:
Na₂O₂ + CO₂ → Na₂CO₃ + 1/2 O₂
From the stoichiometry of the reaction, 1 mol of Na₂O₂ reacts with 1 mol CO₂. So, the stoichiometric ratio is 1 mol Na₂O₂/1 CO₂.
Now, we convert the mass of reactants to moles by using the molecular weight (Mw) of each compound:
Mw (Na₂O₂) = (23 g/mol x 2) + (16 g/mol x 2) = 78 g/mol
moles Na₂O₂ = 96.7 g/(78 g/mol) = 1.24 mol Na₂O₂
Mw (CO₂) = 12 g/mol + (16 g/mol x 2) = 44 g/mol
moles CO₂ = 0.0755 g/(44 g/mol) = 1.71 x 10⁻³ mol CO₂
Now, we calculate the number of moles of CO₂ we need to completely react with the mass of Na₂O₂ we have:
1.24 mol Na₂O₂ x 1 mol CO₂/1 mol Na₂O₂ = 1.24 mol CO₂
In 1 L of respired air we have 1.71 x 10⁻³ mol CO₂ (0.0755 g), so we need the following number of liters to have 1.24 mol of CO₂:
1.24 mol CO₂ x 1 L/1.71 x 10⁻³ mol CO₂ = 725.15 L
