The metabolic oxidation of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) in our bodies produces \(\mathrm{CO}_{2},\) which is expelled from our lungs as a gas: $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ (a) Calculate the volume of dry \(\mathrm{CO}_{2}\) produced at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) and 0.970 atm when \(24.5 \mathrm{~g}\) of glucose is consumed in this reaction. (b) Calculate the volume of oxygen you would need, at 1.00 atm and \(298 \mathrm{~K}\), to completely oxidize \(50.0 \mathrm{~g}\) of glucose.

First, we need to determine the moles of glucose consumed by using its molar mass and given mass in the problem. 1. Convert the mass of glucose (C₆H₁₂O₆) to moles. Moles of glucose = \(\frac{Mass of glucose}{Molar mass of glucose}\) Molar mass of glucose (C₆H₁₂O₆) = \(6 \times 12.01 g/mol + 12 \times 1.01 g/mol + 6 \times 16.00 g/mol\) After calculating the molar mass, we can use the given mass 24.5 g to find the moles of glucose: Moles of glucose = \(\frac{24.5 g}{Molar mass of glucose}\) 2. Using the balanced chemical equation, we can find the moles of CO₂ produced. The stoichiometry of glucose to CO₂ is 1:6, which means that for every one mole of glucose, six moles of CO₂ are produced. Moles of CO₂ = \(6 \times\) Moles of glucose 3. Find the volume of CO₂ produced using the Ideal Gas Law (PV = nRT). In this part, we are given the temperature (37 ºC = 310.15 K) and pressure (0.970 atm). The Ideal Gas Law constant R is 0.0821 L atm/mol K. Volume of CO₂ (V) = \(\frac{nRT}{P}\) V = \(\frac{Moles\: of\: CO₂ \times 0.0821 L\:atm/mol\:K × 310.15 K}{0.970\: atm}\) Now, you can calculate the volume of CO₂ produced.

1. Use the given mass of glucose (50.0 g) to find the moles of glucose. Moles of glucose = \(\frac{50.0 g}{Molar mass of glucose}\) 2. Using the balanced chemical equation, find the moles of O₂ needed. The stoichiometry of glucose to O₂ is 1:6, which means that for every one mole of glucose, six moles of O₂ are required. Moles of O₂ = \(6 \times\) Moles of glucose 3. Find the volume of O₂ needed using the Ideal Gas Law (PV = nRT). In this part, we are given the temperature (298 K) and pressure (1.00 atm). The Ideal Gas Law constant R is 0.0821 L atm/mol K. Volume of O₂ (V) = \(\frac{nRT}{P}\) V = \(\frac{Moles\: of\: O₂ \times 0.0821 L\:atm/mol\:K × 298 K}{1.00\: atm}\) Now, you can calculate the volume of O₂ needed to completely oxidize 50.0 g of glucose.