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Chapter 10: Problem 57

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 normal body temperature, \(37^{\circ} \mathrm{C},\) and \(101.33 \mathrm{kPa}\) when $10.0 \mathrm{~g}$ of glucose is consumed in this reaction. (b) Calculate the volume of oxygen you would need, at \(100 \mathrm{kPa}\) and \(298 \mathrm{~K},\) to completely oxidize \(15.0 \mathrm{~g}\) of glucose.

Short Answer

Expert verified
(a) The volume of dry CO2 produced at normal body temperature (37°C) and pressure (101.33 kPa) when 10.0g of glucose is consumed is 8.36 L. (b) The volume of oxygen required to completely oxidize 15.0g of glucose at 100 kPa and 298 K is 12.39 L.

Step by step solution

01

Convert the temperature to Kelvin

The temperature given for the body is 37°C. We need to convert this to Kelvin by adding 273.15: Temperature in Kelvin = 37°C + 273.15 = 310.15 K
02

Find the moles of glucose used

The mass of glucose (C6H12O6) used is 10.0g. First, let's determine the molar mass of glucose: Molar mass of C6H12O6 = (6 × 12.01) + (12 × 1.01) + (6 × 16.00) = 180.18 g/mol Now, let's find the moles of glucose used: Moles of glucose = (10.0 g) / (180.18 g/mol) = 0.05553 mol
03

Find the moles of CO2 produced

The stoichiometric ratio of glucose to CO2 in the reaction is 1:6. Therefore, for every mole of glucose, 6 moles of CO2 are produced. Moles of CO2 = 0.05553 mol × 6 = 0.3332 mol
04

Apply the ideal gas law for CO2

Now, we can apply the ideal gas law to find the volume of CO2 produced: \(PV = nRT\) where P is the pressure, V is the volume, n is the moles of CO2, R is the ideal gas constant, and T is the temperature in Kelvin. We will use the ideal gas constant value R = 8.314 J/(mol·K), as our pressure is given in kPa. Rearrange the equation to solve for volume: V = nRT / P Substitute the pressure, moles of CO2, ideal gas constant, and temperature into the equation: V = (0.3332 mol × 8.314 J/(mol·K) × 310.15 K) / (101.33 kPa) = 8.36 L Now, following the same approach, we will find the volume of O2 needed for part (b).
05

Find the moles of glucose for part (b)

The mass of glucose used in this part is 15.0g. Moles of glucose = (15.0g) / (180.18g/mol) = 0.0833 mol
06

Find the moles of O2 needed

The stoichiometric ratio of glucose to O2 in the reaction is 1:6. Therefore, for every mole of glucose, 6 moles of O2 are required. Moles of O2 = 0.0833 mol × 6 = 0.4998 mol
07

Apply the ideal gas law for O2

We are given the pressure (100 kPa) and temperature (298 K) for part (b). Apply the ideal gas law to find the volume of O2 needed: V = nRT / P Substitute the pressure, moles of O2, ideal gas constant, and temperature into the equation: V = (0.4998 mol × 8.314 J/(mol·K) × 298 K) / (100 kPa) = 12.39 L So, the volume of dry CO2 produced at normal body temperature and pressure when 10.0g of glucose is consumed is 8.36 L, and the volume of oxygen required to completely oxidize 15.0g of glucose at 100 kPa and 298 K is 12.39 L.

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