Ma Problem 10.57-Enhanced with Feedback )9 of 17 Part A Calculate he volume ofdr

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Ma Problem 10.57-Enhanced with Feedback )9 of 17 Part A Calculate he volume ofdry O, produced at body temperature (37·q and 0 990 atm when 24.5 g of glicose is consumed in this reaction The metabolic axidation of ghicose, CHaOs.in our bodies produces COa. which is expelled from our lungs as a gas Express your answer using three significant figures HaOs (ag) +60a(g)+6C0(9)+6H20(0) You may want to reference (Pages 408-410) Section 10 5 while completing this problem Submt Part B Calculate the volume of oxygen you would need, at 100 atzm and 298 K to completely xidice 52 of glucose Express your answer using three significant figures O Type here to search

Explanation / Answer

C6H12O6(aq) + 6 O2(g) ------------------- 6CO2(g) + 6 H2O(l)

1 mole                                             6 mole

mass of glucose = 24.5 grams

molar mass of glucose = 180.16 gram/mole

number of moles of glucose = 24.5/180.16 =0.136 moles

according to equation

1 mole of glucose = 6 moles of CO2

0.136 molesof glucose = ?            

                             = 0.136x6/1=0.816 moles of CO2

number of moles of CO2= 0.816 moles

P= 0.990 atm

T=37C =37+273= 310 K

at 37C

vapour pressure of water = 0.062 atm

Pressure of Dry gas = 0.990 - 0.062 =0.928 atm

R=0.0821 L-atm/mole-K

PV=nRT

V= nRT/P= 0.816x0.0821x310/0.928 =22.38L

Volume of dry CO2 = 22.38L

2)

P= 1atm

T= 298K

R=0.0821 L-atm/mol-k

mass of glucose= 52 grams

molar mass of glucose = 180.16 gram/mole

number of moles of glucose = 52/180.16 =0.289 moles

according to equation

1 mole of glucsoe = 6 moles of O2

0.289 moles of glucose = ?

                                       = 6x0.289/1= 1.734 moles of O2

number of moles of O2= 1.734 moles

V=nRT/P

V= 1,734x0.0821x298/1.00 =42.42 L

volume of O2 = 42.42L

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