Answer: H₂O₂ (94%) > Air (23%) > Exhaled air (13%) > NaHCO₃ (0%)
Initial important note:
Although NaHCO₃ contents oxygen atoms, and you can calculate its compositoin, the resulting gas does not containg pure oxygen gas (O₂). For the comparisson it is not useful to calculate the content of oxygent atoms, but the concentration of O₂ gas. As such, the gas from NaHCO₃ contains 0% of pure O₂, that is why it is ranked last.
1) Air:
Source: internet
Approximate 23%. It is variable, because air is not a pure substance but a mixture of gases, whose compositon is not unique.
2) Exhaled air:
Source: internet.
Approximate 13%. The compositon of the air changes in our lungs, due to the respiration process: we inhale fresh air with around 23% of oxygen, part of this oxygen pass to the cells (lungs - blood - heart - cells) and then it is exhaled with a lower content of air and a greater content of CO₂
3) Air from the decomposition of H₂O₂.
In this case we can do a chemical calculation, since we can state the chemical equation of the reaction:
i) Chemical Equation:
H₂O₂ (g) → H₂ (g) + O₂ (g)
ii) mole ratio of the products 1 mol H₂ : 1 mol O₂
iii) convert moles into mass (grams)
1 mol H₂ × 2 × 1.008 g/mol = 2.016 g
1 mol O₂ × 2 × 15.999 g/mol = 31.998 g
Composition, % = [31.998 g / (2.016 g + 31.998 g) ] × 100 ≈ 94%
4) Air from the decomposition of NaHCO₃:
i) chemical equation:
2 NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g)
ii) mole ratio: take into account only the gases in the products:
1 mol CO₂ (g) : 1 mol H₂O
iii) mass in grams
CO₂: molar mass ia approximately 44.01 g/mol
H₂O: molar mass is approximately 18.02 g/mol
iii) Those gases although have oxygen atoms, do not hae free oxygen gas, which is what we are compariing. That means, that from the decomposition of NaHCO₃ you get 0% oxygen gas.
5) The result is:
H₂O₂ (94%) > Air (23%) > Exhaled air (13%) > NaHCO₃ (0%)
