15. -15 points My Notes Ask Your T A person inhales air richer in 02 and exhales

Question

15. -15 points My Notes Ask Your T A person inhales air richer in 02 and exhales air richer in CO2 and water vapor. During each hour of sleep, a person exhales a total of 300. L of this CO2-enriched and H20 enriched air. (a) If the partial pressures of CO2 and H20 in exhaled air are each 40.0 torr at 36.2°C, calculate the masses of CO2 and of H20 exhaled in 1.00 hour of sleep. CO2 H20 (b) How many grams of body mass does the person lose in about 8 h sleep if all the CO2 and H20 exhaled come from the metabolism of glucose? (Note: 8 h has 1 significant figure, so round your answer to 1 sig. fig.) C6H12O6(s) 6 02(g) 6 CO2(g) + 6 H2O(g)

Explanation / Answer

According to the Dalton's Law total Pressure:

Ptotal = PCO2 + PH2O
Ptotal = 40.0 Torr + 40.0 Torr = 80.0 Torr

Now to determine the amount of moles with the use of Ideal Gas Equation as follows:

PV = nRT
Lets use R = 62.364 (L*Torr)/(K*mol)

T=36.2 °C= 309.2 K

Therefore;

n = (R*T)/(P*V) = (62.364*309.2 )/(60.0*300L)

= 1.071 total moles

Now use mole fraction to calculate the moles of CO2 and H2O:

xi = Pi/P = ni/n

where:
xi = mole fraction of any individual gas component in a gas mixture
Pi = partial pressure of any individual gas component in a gas mixture
ni = moles of any individual gas component in a gas mixture
n = total moles of the gas mixture
P = total pressure of the gas mixture

nCO2 = n*(PCO2/P) = 1.071*(30.0/60.0)

= 0.5355 mol of CO2
nH2O = n*(PH2O/P) = 1.071*(30.0/60.0)

= 0.5355 mol of H2O

Amount of any substance = molar mass* number o fmoles

CO2: 44.01 g/mol *0.5355 mol of CO2
= 23.57 g

H2O = 18.02 g/ mole*0.5355 mol of CO2
= 9.65 g

MW of CO2 = 44.01 g/mol and H2O = 18.02 g/mol


b) C6H12O6 MW = 180.16 g/mol

0.535 5 mol of CO2 per hour =0.535 5 mol*8

= 4.284 moles of CO2 exhaled in 8 hours.

Using Stoichiometry

C6H1206 (s) + 6O2 (g) ---> 6CO2 (g) + 6H2O (g)

4.284 moles CO2 x (1 mol C6H12O6/6 moles CO2) =

0.714 mole Glucose

Amount of glucose or body mass

= 0.714 mole Glucose * (180.16 g C2H12O6/ 1 mol C2H12O6)

= 128.63 grams of body mass loss

=1.0*10^2 g (one significant )

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