Question

Problem 10.57 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)	Part A Calculate the volume of dry CO2 produced at body temperature (37 ℃) and 0.990 atm when 24.0 g of glucose is consumed in this reaction.

Answer 1

Here we use ideal gas law.

pV = nRT

V = nRT / p

Here p = 0.990 atm, R = 0.08206 L atm per (K mol )

T= 37 + 273.15 = 310.15 K

We must need to find moles of CO2.

In order to get moles of CO2, we use given mass of glucose = 24 g

Calculation of moles of glucose

Moles of glucse = 24.0 g / molar mass of glucse

            = 24 .0 g / 180.1559 g per mol

`           = 0.13321 mol glucose

Moles of CO2

To calculate moles of CO2 we use mole ration between CO2 : Glucose.

1 mol Glucose : 6 mol CO2

Therefore moles of CO2 = 0.13321 mol glucose x 6 mol CO2 / 1 mol glucose

= 0.7993 mol glucose

Calculation of volume of CO2

Volume of CO2 =(0.7993 mol CO2 x 0.08206 L atm ( K mol ) x 310.15 K ) / 0.990 atm

= 20.54 L

So the volume produced of CO2 = 20.54 L