Assume that an exhaled breath of air consists of $74.8 \% \mathrm{~N}_{2}, 15.3 \% \mathrm{O}_{2}, 3.7 \% \mathrm{CO}_{2},\( and \)6.2 \%$ water vapor. (a) If the total pressure of the gases is \(99.8 \mathrm{kPa}\), calculate the partial pressure of each component of the mixture. (b) If the volume of the exhaled gas is \(455 \mathrm{~mL}\) and its temperature is \(37^{\circ} \mathrm{C},\) calculate the number of moles of \(\mathrm{CO}_{2}\) exhaled. (c) How many grams of glucose ( $\left.\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)$ would need to be metabolized to produce this quantity of \(\mathrm{CO}_{2} ?\) (The chemical reaction is the same as that for combustion of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\). See Section 3.2 and Problem \(10.57 .\) )

We are given the percentage composition of the exhaled breath of air and its total pressure (99.8 kPa). We will use this information to calculate the partial pressure of each gas component. The formula for partial pressure is: Partial Pressure = (% of gas component) × (Total Pressure) So, for each gas component, we will calculate the partial pressure as follows: 1) N₂: Partial Pressure (N₂) = \(0.748 \times 99.8 \mathrm{kPa} = 74.7 \mathrm{kPa}\) 2) O₂: Partial Pressure (O₂) = \(0.153 \times 99.8 \mathrm{kPa} = 15.3 \mathrm{kPa}\) 3) CO₂: Partial Pressure (CO₂) = \(0.037 \times 99.8 \mathrm{kPa} = 3.69 \mathrm{kPa}\) 4) H₂O: Partial Pressure (H₂O) = \(0.062 \times 99.8 \mathrm{kPa} = 6.19 \mathrm{kPa}\)

We are given the volume of the exhaled gas (455 mL) and its temperature (37°C). We can use this information along with the partial pressure of CO₂ (calculated in Step 1) to find the number of moles of CO₂ exhaled by using the ideal gas law: PV = nRT Where: P = partial pressure of CO₂ (3.69 kPa) V = volume (0.455 L) n = number of moles R = gas constant (8.314 J mol⁻¹K⁻¹ or 8.314 kPa L mol⁻¹K⁻¹) T = temperature in Kelvin (37°C + 273.15 = 310.15 K) Rearranging the formula to find the number of moles (n): n = PV / RT n = \((3.69 \mathrm{kPa})(0.455 \mathrm{L})/((8.314 \mathrm{kPa\:L\:mol}^{-1} \mathrm{K}^{-1})(310.15\: \mathrm{K}))=0.00217\: \mathrm{mol}\) The number of moles of CO₂ exhaled is 0.00217 moles.

To find out how much glucose is metabolized to produce the calculated amount of CO₂, we will use the stoichiometry of the chemical reaction: C₆H₁₂O₆ (glucose) + 6 O₂ → 6 CO₂ + 6 H₂O From the balanced equation, we can see that 1 mole of glucose produces 6 moles of CO₂. So, we can calculate the moles of glucose needed to produce 0.00217 moles of CO₂: Moles of glucose = \((0.00217 \mathrm{mol\: CO_{2}})(1 \mathrm{mol\: C_{6} H_{12} O_{6}}/6\: \mathrm{mol\: CO_{2}})= 3.62 \times 10^{-4}\: \mathrm{mol\: C_{6} H_{12} O_{6}}\) Next, we will convert the moles of glucose to grams using the molar mass of glucose, which is 180.16 g/mol: Mass of glucose = \((3.62 \times 10^{-4}\: \mathrm{mol\: C_{6} H_{12} O_{6}})(180.16 \mathrm{g/mol})=0.0652\: \mathrm{g}\) Therefore, approximately 0.0652 grams of glucose needs to be metabolized to produce the calculated amount of CO₂ exhaled.