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 degree C) and 0.990 atm when 24.0 g of glucose is consumed in this reaction.
2.) Calculate the volume of oxygen you would need, at 1.00 atm and 298 K, to completely oxidize 55g of glucose.

Answer 1

1) 20.5 L CO2

2) 44.75 L O2

Step-by-step explanation:

Step 1: Data given

Temperature = 37.0 °C

Pressure = 0.990 atm

Mass of glucose = 24.0 grams

Molar mass glucose: 180.156 g/mol

Step 2: The balanced equation

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

Step 3: Calculate moles glucose

Moles glucose = mass glucose / molar mass glucose

Moles glucose = 24.0 grams / 180.156 g/mol

Moles glucose = 0.133 moles

Step 4: Calculate moles CO2

For 1 mol glucose we need6 moles O2 to produce 6 moles CO2 and 6 moles H2O

For 0.133 moles glucose we'll have 6*0.133 = 0.798 moles CO2

Step 5: Calculate volume CO2

p*V = n*R*T

⇒ p = the pressure = 0.990 atm

⇒ V = the volume of CO2 = TO BE DETERMINED

⇒ n = the number of moles CO2 = 0.798 moles

⇒ R = the gas constant = 0.08206 L* atm / K*mol

⇒ T = the temperature = 37.0 °C = 310 K

V = (nRT)/p

V = (0.798 * 0.08206 * 310) / 0.990

V= 20.5 L CO2

2.) Calculate the volume of oxygen you would need, at 1.00 atm and 298 K, to completely oxidize 55g of glucose.

Calculate moles glucose

Moles glucose = mass glucose / molar mass glucose

Moles glucose = 55.0 grams / 180.156 g/mol

Moles glucose = 0.305 moles

Calculate moles O2

For 1 mol glucose we need6 moles O2 to produce 6 moles CO2 and 6 moles H2O

For 0.305 moles we need 6 * 0.305 = 1.83 moles O2

Calculate volume of O2

p*V = n*R*T

⇒ p = the pressure = 1.00 atm

⇒ V = the volume of O2 = TO BE DETERMINED

⇒ n = the number of moles O2 = 1.83 moles

⇒ R = the gas constant = 0.08206 L* atm / K*mol

⇒ T = the temperature = 298 K

V = (nRT)/p

V = (1.83 * 0.08206 * 298) / 1.00

V= 44.75 L O2

Answer 2

1. 20.5 L of CO₂ are released.

2. 44.8L of O₂ are needed

Step-by-step explanation:

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

We assume oxygen as reagent in excess.

Let's convert the mass of glucose to moles → 24 g . 1mol/180 g = 0.133 moles

Ratio is 1:6 so (0.133 . 6) = 0.8 moles of CO₂ are released

We apply the Ideal Gases Law to determine the volume:

P . V = n . R .T → V = (n. R .T)/P

V = (0.8 mol . 0.082 . 310K) /0.990 atm = 20.5L

For 2nd question: First of all, we determine the moles of glucose

55 g . 1mol/180 g = 0.305 moles

We apply a rule of three: 1 mol of glucose needs 6 moles of O₂ to react

0.305 moles may need (0.305 .6)/1 = 1.83 moles

We use the Ideal Gases Law again:

V = (1.83 mol . 0.082 . 298K) / 1 atm = 44.8L