Question

A person inhales air richer in O2 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 H2O-enriched air. (a) If the partial pressures of CO2 and H2O in exhaled air are each 20.0 torr at 26.5°C, calculate the masses of CO2 and of H2O exhaled in 1.00 hour of sleep. CO2 ___ ?g

H2O_____ ?g

(b) How many grams of body mass does the person lose in about 8 h sleep if all the CO2 and H2O exhaled come from the metabolism of glucose? (Note: 8 h has 1 significant figure, so round your answer to 1 sig. fig.) ____? g

C6H12O6(s) + 6 O2(g) ---> 6 CO2(g) + 6 H2O(g)

Answer 1

partial pressure = mole fraction x total pressure

let's do a little derivation....

PCO2 = (moles CO2 / moles total) x Pt
moles CO2 = (PCO2 / Pt) x moles total

since moles = mass / molar mass
mass = moles x molar mass

if we multiply both sides by molar mass CO2

moles CO2 x molar mass CO2 = (PCO2 / Pt) x moles total x molar mass CO2
and substitute
mass CO2 = (PCO2 / Pt) x moles total x molar mass CO2

next let's use the ideal gas law to find "moles total"...from
PV = nRT
n = PV/RT = (Pt atm) x (300 L) / [(0.0821 Latm/moleK) x (299.5K)]
= 12.2 (moles/atm) x Pt (atm)

substituting...

mass CO2 = (PCO2 / Pt) x (12.20 (moles/atm) x Pt (atm)) x (44.0 g/mole)
or
mass CO2 = PCO2 (in atm) x 12.20 (moles/atm) x 44.0 (g/mole)
for H2O
mass H2O = PH2O (in atm) x 12.20 (moles/atm) x 18.0 (g/mole)

finally...

PCO2 = PH2O = 20.0 torr x (1 atm / 760 torr) = 0.0263 atm
and this gives

mass CO2 = (0.0263 atm) x 12.2 (moles/atm) x 44.0 (g/mole) = 14.1g
mass H2O = (0.0263 atm) x 12.2 (moles/atm) x 18.0 (g/mole) = 5.78g

both have 3 sig figs
************************

C6H12O6(s) + 6 O2(g) -----> 6 CO2(g) + 6 H2O(g)

moles CO2 = 14.12 / 44.0 = 0.3209 moles / hr
moles H2O = 5.78/18.0 = 0.321 moles / hr

either is the limiting product but let's base our calculations on CO2

from balanced equation 1 mole glucose ----> 6 moles CO2

8 hr x (0.321 moles CO2 / hr) x (1 mole glucose / 6 moles CO2) x (180 g/mole) = 77.067 g = 8 x10^1 g with 1 sig fig

so the person drops 8x10^1 g of body mass in an 8 hr period