Answer:
3.11 L is the volume for the produced CO₂
Explanation:
The metabolic breakdown of glucose is the respiration reaction. It's a redox reaction type. The equation, as shown is:
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)
We assume that the oxygen is in excess, so the glucose is the limiting reagent. We define the moles:
3.68 g . 1mol / 180g = 0.0204 moles
Now we propose this rule of three:
1 mol of glucose produces 6 moles of CO₂
Therefore, 0.0204 moles of glucose may produce (0.0204 . 6 ) / 1 = 0.1224 moles of glucose. Now we apply the Ideal Gases Law, to find out the volume: P . V = n . R . T
V = (n . R . T) / P
37°C + 273 = 310K
V = (0.1224 mol . 0.082 L.atm/mol.K . 310K) / 1 atm = 3.11 L
Considering the reaction stoichiometry and ideal gas law, the volume of CO₂ produced at 37°C and 1.00 atm when 3.68 g of glucose is used up in the reaction is 3.11 L.
The balanced reaction is:
C₆H₁₂O₆(s) + 6 O₂(g) → 6 CO₂(g) + 6 H₂O(l)
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
Being the molar mass of glucose 180 g/mole, if 3.68 grams react, the number of moles that react is calculated as:
3.68 grams×[tex]\frac{1 mole}{180 grams}[/tex]= 0.0204 moles
Then you can apply the following rule of three: if by stoichiometry 1 mole of C₆H₁₂O₆ produce 6 moles of CO₂, 0.0204 moles of C₆H₁₂O₆ produce how many moles of CO₂?
[tex]amount of moles of CO_{2} =\frac{0.0204 moles of C_{6} H_{12} O_{6} x6moles of CO_{2} }{1mole of C_{6} H_{12} O_{6}}[/tex]
amount of moles of CO₂= 0.1224 moles
On the other side, an ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:
P× V = n× R× T
In this case, for CO₂, you know:
Replacing in the ideal gas law:
1 atm× V = 0.1224 moles× 0.082 [tex]\frac{atmL}{molK}[/tex]× 310 K
Solving:
[tex]V=\frac{0.1224 molesx 0.082\frac{atmL}{molK}x310 K}{1 atm}[/tex]
V=3.11 L
Finally, the volume of CO₂ produced at 37°C and 1.00 atm when 3.68 g of glucose is used up in the reaction is 3.11 L.
Learn more:
