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 about 300. L of this CO₂−enriched and

H₂O-enriched air.

(a) If the partial pressures of CO₂ and H₂O in exhaled air are each 29.3 torr at 37.0 C, calculate the masses of CO₂ and of H₂O exhaled in 1 h of sleep

____g CO2

____g H2O

(b) How many grams of body mass does the person lose in 8 h sleep if all the CO₂ and H₂O exhaled come from the metabolism of glucose? Assume that the mass lost is equal to the mass of glucose that is converted to CO₂ and H₂O.

C₆H₁₂O₆(s) + 6 O₂(g) → 6 CO₂(g) + 6 H₂O(g)

___g

Answer 1

a) Using ideal gas equation

PV = nRT

P = 29.3 torr = 29.3/760 atm

V = 300 L

Temperature = 273 + 37.0 = 310.0 K

29.3/760 * 300 = n * 0.0821 * 310.0

n = 0.4544 moles

Mass of CO2 = number of moles * molar mass = 0.4544 mol * 44 g/mol = 19.99 grams

Mass of H2O = number of moles * molar mass = 0.4544 mol * 18 g/mol = 8.18 grams

(b)

As per the balanced reaction, 1 mole of glucose will form 6 moles of CO2 and 6 moles of H2O

Moles of glucose lost = (0.4544)/6 = 0.075733 moles

Molar mass of glucose (C6H12O6) = 180.15 g/mol

Mass lost = number of moles * molar mass = 0.075733 mol * 180.15 g/mol = 13.64 grams

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