The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas:
C6H12O6(aq)+6O2(g)→6CO2(g)+6H2O(l)
Calculate the volume of dry CO2 produced at body temperature (37 ∘C) and 0.990 atm when 24.0 g of glucose is consumed in this reaction.
use 3 sig figs
plz show your work
Cody Jinks is almost right. Everything Cody writes is OK except the problem asks for DRY CO2 and since this is produced from the body it will be wet CO2; therefore, instead of using 0.990 atm for pressure use 0.990 - vapor pressure H2O at 37 C You can look up that value on the web or in vapor pressure tables in almost any chemistry book.
Use the balanced equation to convert 24g of glucose to mol of glucose then to mol of CO2. Then use the PV=nRT equation to solve for volume.
Ps I might be wrong
To calculate the volume of dry CO2 produced at body temperature and pressure when 24.0 g of glucose is consumed in the given reaction, we need to use the ideal gas law.
The ideal gas law is given by:
PV = nRT
Where:
P = pressure = 0.990 atm
V = volume (in liters)
n = moles of gas
R = ideal gas constant = 0.0821 L·atm/(mol·K)
T = temperature (in Kelvin) = 37 + 273.15 = 310.15 K
We need to find the moles of CO2 produced by converting the mass of glucose into moles using its molar mass and the balanced chemical equation.
Molar mass of C6H12O6:
(6 × atomic mass of C) + (12 × atomic mass of H) + (6 × atomic mass of O)
= (6 × 12.01 g/mol) + (12 × 1.01 g/mol) + (6 × 16.00 g/mol)
= 180.18 g/mol
Moles of glucose (C6H12O6):
Moles = mass / molar mass
= 24.0 g / 180.18 g/mol
≈ 0.133 moles
From the balanced chemical equation, we can see that for every mole of glucose consumed, 6 moles of CO2 are produced. Therefore, 0.133 moles of glucose will produce (6 × 0.133) = 0.798 moles of CO2.
Now we can use the ideal gas law equation to find the volume.
PV = nRT
V = (nRT) / P
V = (0.798 mol × 0.0821 L·atm/(mol·K) × 310.15 K) / 0.990 atm
V ≈ 20.3 L
Therefore, approximately 20.3 liters of dry CO2 will be produced at body temperature (37 °C) and a pressure of 0.990 atm when 24.0 g of glucose is consumed in this reaction.