How much air does Sarah inhale while running?

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SUMMARY

The discussion centers on calculating the volume of air Sarah inhales while running, based on her glucose intake from two oranges, which provide 25 grams of glucose. Using the ideal gas law and the molar mass of glucose (180.18 g/mol), it is determined that Sarah inhales approximately 20.31 liters of oxygen and produces about 1.02 liters each of carbon dioxide and water vapor. The conversation also addresses the concept of limiting reagents, clarifying that glucose acts as the limiting reagent in this metabolic process, while only 5% of the inhaled oxygen is utilized for energy production.

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  • Understanding of the ideal gas law and its application in calculations
  • Knowledge of molar mass and how to convert grams to moles
  • Familiarity with the concept of limiting reagents in chemical reactions
  • Basic principles of aerobic respiration and gas exchange in humans
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This discussion is beneficial for biology students, chemistry enthusiasts, and fitness professionals interested in understanding the metabolic processes involved in exercise and energy production.

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Homework Statement



(The chemical equation is attached)

Problem:

Just before running, Sarah eats two oranges. Oranges provide her body with 25 grams of glucose which allows it to produce energy. The external temperature is 27 C, and atmospheric pressure is 102.3 kPa. While 21% of the air Sarah inhales is composed of O2, she exhales approximately 16% of the oxygen, In other words, she only uses 5% of the oxygen she inhaled.

1. How many litres of air does Tala inhale as she runs to burn the glucose consumed?

2. How many litres of water vapour does she produce?

3. How many litres of carbon dioxide does she produced?

2. The attempt at a solution

Part 1

25 grams of glucose = (1250/9009) moles of glucose.

Oxygen is 6 times the moles of glucose, thus, (7500/9009) moles of O2

Determining the litres of air:

(102.3)(V)=(7500/9009)(8.314)(27+273.15)

V = 20.31 Litres of O2

Part 2 & 3

25 grams of glucose = (1250/9009) moles of glucose.

Oxygen is 6 times the moles of glucose, but only 5% of it is used, therefore:

((1250 * 6) / (9009)) * 0.05 = 125/3003 moles of O2 are used.

This is now the limiting reactant.

(102.3)(v)=(125/3003)(8.314)(27+273.15)

V = 1.02 litres of CO2 and H2O each.

Note: When they stated "oranges", I assumed that both of the oranges provided 25 grams of glucose in total.
 

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It's not clear how 25 g of glucose is equal to (1250/9009) moles of glucose. The MW of glucose is about 180.
 
SteamKing said:
It's not clear how 25 g of glucose is equal to (1250/9009) moles of glucose. The MW of glucose is about 180.

The molar mass of glucose according to the periodic table I have is 180.18 g/mol. It works out as a fraction to (1250/9009) moles.
 
Speedking96 said:
Determining the litres of air:

(102.3)(V)=(7500/9009)(8.314)(27+273.15)

V = 20.31 Litres of O2

Air, or oxygen?

Oxygen is 6 times the moles of glucose, but only 5% of it is used, therefore:

((1250 * 6) / (9009)) * 0.05 = 125/3003 moles of O2 are used.

This is now the limiting reactant.

(102.3)(v)=(125/3003)(8.314)(27+273.15)

V = 1.02 litres of CO2 and H2O each.

No idea what you mean by limiting reagent in this context, but in general, if there are 6 moles of CO2 produced per 6 moles of O2 consumed (see the reaction equation), volume of CO2 must be similar to that of volume of oxygen.

Question about water vapor doesn't make much sense to me, as at these conditions water is liquid.
 
Borek said:
Air, or oxygen?
No idea what you mean by limiting reagent in this context, but in general, if there are 6 moles of CO2 produced per 6 moles of O2 consumed (see the reaction equation), volume of CO2 must be similar to that of volume of oxygen.

Question about water vapor doesn't make much sense to me, as at these conditions water is liquid.

Well, the question itself is quite ambiguous and uses the terms interchangeably. So, I guess it would be correct to say oxygen.

For the limiting reagent; the problem states that only 5% of the actual oxygen that is inhaled is useful, because the rest is exhaled. I understood this as: 5% of the oxygen will play a role in how the products are formed, and thus, it acts as a limiting reagent.
 
Last edited:
Borek said:
Question about water vapor doesn't make much sense to me, as at these conditions water is liquid.

I think the question is asking how much water vapor is produced from metabolizing the glucose in the runner's body. Unless the girl has a runny nose, she will exhale water vapor along with CO2.
 
SteamKing said:
I think the question is asking how much water vapor is produced from metabolizing the glucose in the runner's body. Unless the girl has a runny nose, she will exhale water vapor along with CO2.

Then it should ask how much water is produced, not water vapor.

She will also sweat and pee (probably not while running :rolleyes:).
 
Speedking96 said:
For the limiting reagent; the problem states that only 5% of the actual oxygen that is inhaled is useful, because the rest is exhaled. I understood this as: 5% of the oxygen will play a role in how the products are formed, and thus, it acts as a limiting reagent.

No, that's not what is understood by limiting reagent in chemistry.

Actually I would say glucose is a limiting reagent here - once it ends, no matter how oxygen is present, no more carbon dioxide will be produced.

5% oxygen information is necessary for calculation of the volume of air required from known volume of oxygen that reacted.
 
Borek said:
No, that's not what is understood by limiting reagent in chemistry.

Actually I would say glucose is a limiting reagent here - once it ends, no matter how oxygen is present, no more carbon dioxide will be produced.

5% oxygen information is necessary for calculation of the volume of air required from known volume of oxygen that reacted.

Yes, I agree.

But, wouldn't the amount of "useful" oxygen play a role in how much glucose is actually *used* up. The problem says that she inhales 21% of the oxygen in the air, exhales 16% of the oxygen in the air, and only uses 5%. Perhaps I'm misinterpreting the word *uses*. From what I understand - or maybe not - is that the 16% of oxygen exhaled does not play a role in the reaction as it is simply exhaled, and thus, 5% should be considered. It is weirdly worded.
 
  • #10
No, glucose is ALL used up. There is nothing strange in the wording, at least I don' see anything like that.
 
  • #11
Borek said:
No, glucose is ALL used up. There is nothing strange in the wording, at least I don' see anything like that.
So, what you are saying is that Oxygen does not play any kind of limiting role. What I fail to understand then, is that why do they say that 5% of oxygen is useful? Wouldn't that information be unnecessary as they already say that 21% of the air that is inhaled is Oxygen. Why would they elaborate on what percentage is useful?
 
  • #12
Because that's important for calculating volume of air used.

You can easily calculate how much oxygen was used, but that's not equivalent to the volume of air. The less oxygen of the inhaled air you use, the larger volume of air you need to breathe.

Imagine you calculated you need 1 L of oxygen. This 1 L of oxygen is in almost exactly 5 L of air, so 5 L of air would be sufficient. However, if you are capable of using only half of the oxygen present in the air, you would need 10 L of air.
 

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