Question

The human body gets energy through the slow combustion of glucose (C6H12O6)
the heart obtained by breathing
_____C6H12O6 + _____ O2 -->_____ CO2 + _____ H2O
a) Balance the equation
b) If a person can breathe on average 7.5 L / min of air, what amount of glucose (kg) can
burn in one hour? Consider the average air density as 1.29 kg / m3
c) Get one person's exhale per day if you burned 200 g of glucose.
Consider that the air only contains O2 and N2.

Answer 1

Part a

Balanced chemical reaction

C6H12O6 + 6 O2 --> 6 CO2 + 6 H2O

Part b

Mass of air taken = volumetric flow x density

= 7.5 L/min x 1m3/1000L x 1.29 kg/m3

= 0.009675 kg/min

Moles of air = 0.009675 kg/min / 29 kg/kmol

= 0.0003336 kmol/min

Air consists 21% O2

Moles of O2 = 0.0003336 x 0.21 = 7 x 10^-5 kmol/min

From the stoichiometry of the reaction

6 kmol O2 required to burn = 1 kmol of Glucose

7 x 10^-5 kmol O2 required to burn = 7*10^-5/6

= 1.167 x 10^-5 kmol of Glucose

Mass of glucose burn = 1.167 x 10^-5 kmol x 180.156 kg/kmol

= 0.0021036 kg/min x 60min/hr

= 0.126 kg/hr

Part C

Assume 200g/day glucose burn

Moles of glucose burn = 200 g / 180.156g/mol

= 1.11 mol

O2 exhaled = 1.11*6 = 6.66 mol

Air exhaled = 6.66/0.21 = 31.72 mol

Mass of air exhaled = 31.72 mol x 29 g/mol

= 919.84 g