Gas Laws: C6H12O6+6O2-->6CO2+6H2?

Question

Oxidation of glucose by the body produces carbon dioxide which is expelled through the lungs..

A) Calculate the volume of Co2 produced at body temperature (37 degrees celcius) and 0.970 atm when 24.5 g of glucose is consumed in this reaction?

Answer 1

Let glucose be called G. Mol. wt. G = 180
R = 0.0820 L-atm/mol-K

24.5gG x 1molG/180gG x 6molCO2/1molG = 0.817 mol CO2

PV = nRT

V = nRT/P

V = (0.817mol)(0.0820)(310K)/(0.970atm) = 21.4L CO2