# What amount of forrest-fire in square km or metric tons, would be necesary to bring oxygen levels below 19.5% that humans require to stay alive? [closed]

I read that oxygen makes up 21% of the Earth's atmosphere. I'm guessing it would take a lot of burning to decrease that level, but I am not sure how to calculate that.
The problem is: Roughly 400 hundred years in the future, humans are thinking they need to burn some forests made predominantly of a plant species that produces a gas toxic to humans and other life-forms, that spread on Earth and suffocated a lot of other useful (for us) plant species. They were worried that the large-scale fires would lower the amount of oxygen in the atmosphere to a level that humans would suffocate, but I can tell from the existing answer, that’s not going to be a problem, so they can go ahead with the plan.
I got the 19.5% threshold from this article: https://sciencing.com/minimum-oxygen-concentration-human-breathing-15546.html But on a closer reading, I noticed it said the OPTIMAL range is between 19.5% and 23.5%, and the CRITICAL threshold for survival is 6%, so my mistake. Again, this makes the worries of the people in my story unfounded. My question has been answered, thank you!

## closed as unclear what you're asking by Mołot, Ash, Renan, James♦Nov 14 at 16:58

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• Please clarify your question: Do you literally want to set the (fictional) world on fire to bind enough oxygen to: (A) lower the amount of oxygen in the atmosphere to a level that humans would suffocate or (B) Change the composition of the atmosphere so that oxygen makes up 19,5% of the volume? Please keep in mind that oxygen content depends on the altitude (people on Mt. Everest suffocate even in our current atmosphere). – Elmy Nov 14 at 7:24
• Are you saying people would suffocate if oxygen drops below 19.5 volume percent? That sounds extremely strange. Do you have any source for it? – mathreadler Nov 14 at 11:23
• See also, here: earthscience.stackexchange.com/questions/8930/… – kingledion Nov 14 at 11:56
• I don't know where you got that 19.5% figure from but it's way off. I've done some pretty tough cycling at an altitude equivalent to around 15% at sea level -- it was certainly hard but people live higher than that. Here's a nice paper Hypsographic demography: The distribution of human population by altitude, Joel E. Cohen and Christopher Small, PNAS November 24, 1998 95 (24) 14009-14014 showing that around 20% of people have less than 19.5% O2 sea level equivalent due to altitude – Chris H Nov 14 at 13:27
• ... effectively you're just lifting everyone by about 600m (the source @mathreadler asked for probably doesn't exist, in other words). Some people live above 4000m, which is equivalent to 12% O2 at sea level – Chris H Nov 14 at 13:33

Let's start with considering the entire mass of the atmosphere:

The total mean mass of the atmosphere is $$5.1480 \cdot 10^{18} kg$$.

We know that Oxygen accounts for 21% in volume, and considering that

The density of air at sea level is about $$1.2 \ kg/m^3$$

We get that the "sea level" volume of the entire atmosphere is

$$V = 5.1480 \cdot 10^{18} \over 1.2=4.29 \cdot 10^{18} m^3$$.

You want to consume 1.5% of that volume (21 - 19.5), thus you want to consume

$$M_{O_2}=4.29 \cdot10^{18} [m^3]\cdot 0.015 \cdot 1.429 [{kg\over m^3}]= 9.19 \cdot 10^{16} kg$$ of Oxygen, considering that Oxygen density is $$1.429 [{kg\over m^3}]$$.

Assuming you want to burn Carbon to consume all that Oxygen, how much Carbon would you need?

The chemical reaction for Carbon oxidation is

$$C + O_2 = CO_2 + heat$$

therefore for each mole of Oxygen you need a mole of Carbon. Considering that Oxygen to Carbon atomic weight ratio is 32/12, you would need

$$M_C = M_{O_2} \cdot 12 \over 32$$ $$=9.19 \cdot 10^{16} \cdot 12\over 32=3.5 \cdot 10^{16} kg$$ of Carbon.

The 2011 estimated coal reserves in the entire world amount to $$891 \cdot 10^{12} \ kg$$, just to give you a reference.

As additional note, burning that much carbon would release (taking the heat of combustion of anthracite)

$$Heat = 32 [MJ/kg] \cdot 891 \cdot 10^{12} \ [kg] = 28.5 \cdot 10^{15} \ MJ$$, corresponding to about $$6.7 \cdot 10^6 MTon$$, or about 1 million Tsar bombs...

• Beat me to it. According to this website, the total mass of carbon based life is 5,5e14 kg. So.… pretty hard to do that. vox.com/science-and-health/2018/5/29/17386112/… – Jens Nov 14 at 7:37
• On top of that as atmospheric levels dropped the oceans would give up dissolved oxygen back to the atmosphere as well, there'a about 970 million cubic kilometers of O2 down there. – Ash Nov 14 at 10:17
• @Jens you got me wondering. Given that there was no oxygen in atmosphere until life freed it from co2, and some oxygen sank into iron oxide before atmosphere started to accumulate o2 — where did all that carbon go? We should have more than enough to bring atmospheric levels to 0, and it looks like we don't have enough to get it down by mere percents. – Mołot Nov 14 at 12:28
• @Mołot: Answer: the Oxygen Catastrophe was followed by limestone precipitation which removed the excess carbon from the biosphere. – Joshua Nov 14 at 16:20
• @L.Dutch: Yeah, and the O3 came from water releasing H which is eventually lost to space. – Joshua Nov 14 at 16:36