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

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

Step-by-step 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)* \frac{\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) * \frac{\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} * V & = & \text{0.8159 mol} * \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^(-1)\text{mol}^(-1) * \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}`