Question

The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas:
C6H12O6(aq) + 6 O2(g) = 6 CO2(g) + 6 H2O(l)
(a) Calculate the volume of dry CO2 produced at body temperature (37°C) and 0.970 atm when 24.5 g of glucose is consumed in this reaction. (b) Calculate the volume of oxygen you would need, at 1.00 atm and 298 K, to completely oxidize 50.0 g of glucose.

Answer 1

(a) T = 37 + 273.15 = 310.15 K

p = 0.97 atm

m(gluc) = 24.5 g

M(gluc) = 180 g/mol

n(gluc) $= \frac{24.5}{180} = 0.136 \;mol$

n(CO₂) = 6n(gluc) $= 6 \times 0.136 = 0.816 \;mol$

Ideal gas law: pV = nRT

R = 0.08206 L×atm/mol×K

V $= \frac{nRT}{p}$

V(CO₂) $= \frac{0.816 \times 0.08206 \times 310.15}{0.970} = 21.41 \;L$

(b) T = 298 K

p = 1.0 atm

m(gluc) = 50 g

n(gluc) $= \frac{50}{180} = 0.277 \;mol$

n(O₂) = 6n(gluc) $= 6 \times 0.277 = 1.666 \;mol$

V(O₂) $= \frac{1.666 \times 0.08206 \times 298}{1.00} = 40.75 \;L$