Question

The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas.
C6H12O6(aq) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)

Calculate the volume of dry CO2 produced at body temperature (37°C) and 0.960 atm when 24.5 g of glucose is consumed in this reaction.

Answer 1

Answer:

$\large \boxed{\text{21.6 L}}$

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:        180.16

         C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g:      24.5

(a) Moles of C₆H₁₂O₆

$\text{Moles of C _{6} H _{12} O}_{6} = \text{24.5 g C _{6} H _{12} O}_{6}\times \dfrac{\text{1 mol C _{6} H _{12} O}_{6}}{\text{180.16 g C _{6} H _{12} O}_{6}}\\\\= \text{0.1360 mol C _{6} H _{12} O}_{6}$

(b) Moles of CO₂

$\text{Moles of CO}_{2} =\text{0.1360 mol C _{6} H _{12} O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C _{6} H _{12} O}_{6}} = \text{0.8159 mol CO}_{2}$

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37  °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

$\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is \large \boxed{\textbf{21.6 L}} }$