Question

In: Chemistry

The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our...

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.

Solutions

Expert Solution


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


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