Question

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)

1. Calculate the volume of dry CO2 produced at body temperature (37 ∘C) and 0.960 atm when 23.5 g of glucose is consumed in this reaction. Express your answer using three significant figures in Liters

2. Calculate the volume of oxygen you would need, at 1.00 atm and 298 K, to completely oxidize 54 g of glucose. Express your answer using three significant figures using liters

Answer 1

1. C6H12O6 + 6 O2 = 6 CO2 + 6 H2O

Temperature = 37°C = 310.15 K ,

Moles of glucose = mass / molar mass

= 23.5 / 180

= 0.13056

Moles of CO2 = 6* moles of glucose

= 6*0.13056

= 0.7834

Volume = moles * gas constant * temperature / pressure

= (0.7834 * 0.082 * 310.15 ) / 0.960

= 20.75 L or 20.8 L

2. Moles of glucose = 54 / 180 = 0.3

Moles of O2 = 6 * moles of glucose

= 6 * 0.3 = 1.8

Volume = nRT / P

= (1.8*0.082*298) / 1.00

= 43.985 L

44.0 L

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