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.

1. 20.5 L of CO₂ are released.

2. 44.8L of O₂ are needed

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

1) 20.5 L CO2

2) 44.75 L O2

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

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.

i'm pretty sure its b i might be wrong though!