Question

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 25.5 g of glucose is consumed in this reaction.

Calculate the volume of oxygen you would need, at 1.00 atm and 298 K, to completely oxidize 51 g of glucose.

Answer 1

C6H12O6(aq) + 6O2(g) → 6CO2(g) + 6H2O(l)
1 mole of C6H12O6 produces 6 moles of CO2
25.5/180 moles of C6H12O6 produces 6*25.5/180 moles of CO2
                                  = 0.85
PV = nRT
0.99 * V = 0.85 *0.0821 *(37+273)
V = volume of CO2 = 21.852 L
1mole of C6H12O6 need 6 moles of O2
51/180 moles of C6H12O6 needs 6*51/180 moles of O2
                           = 1.7 moles
PV = nRT
1 * V = 1.7*0.0821*298
V = volume of oxygwn = 41.6 L