The equation for the metabolic breakdown of glucose (C6H12O6) is the same as the equation for the combustion of glucose in air: C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (l) Calculate the volume of CO2 produced at 37°C and 1.00 atm when 3.68 g of glucose is used up in the reaction.

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