Problem 10.57 The metabolic oxidation of glucose, C6H12O6, in our bodies produce

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 C) and 0.990 atm when 24.0 g of glucose is consumed in this reaction.

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 C) and 0.990 atm when 24.0 g of glucose is consumed in this reaction.

Explanation / Answer

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

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