How Many Liters of CO2 Does Eating a Giant Gummy Bear Produce?

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SUMMARY

The discussion centers on calculating the liters of CO2 produced from consuming a giant gummy bear, which contains 1479 grams of glucose. The cellular respiration equation, C6H12O6 + 6 O2 → 6 H2O + 6 CO2 + energy, is utilized to determine the conversion from grams of glucose to moles of CO2. Participants confirm that using stoichiometry and the ideal gas law (PV=nRT) is the correct approach, with body temperature as the assumed temperature and 1 atm as the pressure. This method will yield the accurate volume of CO2 exhaled after consuming the gummy bear.

PREREQUISITES
  • Understanding of cellular respiration and its chemical equation.
  • Knowledge of stoichiometry for converting grams to moles.
  • Familiarity with the ideal gas law (PV=nRT).
  • Basic concepts of thermodynamics, specifically relating to temperature and pressure.
NEXT STEPS
  • Study the stoichiometric conversion of glucose to CO2 in detail.
  • Learn how to apply the ideal gas law (PV=nRT) in biological contexts.
  • Research the physiological aspects of gas exchange in humans.
  • Explore the implications of carbohydrate metabolism on respiratory output.
USEFUL FOR

Students in biology or chemistry, educators teaching cellular respiration, and anyone interested in the metabolic processes related to carbohydrate consumption.

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Homework Statement


(problem won't be word for word, because much of the info. is to throw people off); This large gummy bear, which contains 51 servings and each serving contains 29 grams of carbs. Assuming all the carbs are glucose, how many liters of CO2 would eating this gummy bear cause one to exhale.

Homework Equations


Not a specific equation I'm thinking.

The Attempt at a Solution


So since there is 51 servings with each serving containing 29 grams of carbs, the total amount of carbs which is glucose in this case is 1479 grams. I started out with the equation for cellular respiration:
C6H12O6 + 6 O2 → 6 H2O + 6CO2 + energy. I'm thinking I should take 1479 grams, divide it by the molar mass of glucose, and then convert it to moles of glucose, then moles of CO2. But instead of moles I should treat them as liters instead. Would this get me the right answer?
 
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ngu9997 said:
Would this get me the right answer?
Do you have any specific doubts about this strategy?
 
Yeah I don't feel like I'm doing it right. Would doing all the stoichiometry I stated above to find the moles of CO2 and using PV=nRT work? If so, would the temperature be body temp and would the pressure be 1 atm?
 
ngu9997 said:
Would doing all the stoichiometry I stated above to find the moles of CO2 and using PV=nRT work? If so, would the temperature be body temp and would the pressure be 1 atm?

Yes, yes, yes and yes.
 

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