The metabolic breakdown of glucose (C₆H₁₂O₆) (MW:180.156 g/mol) is given by the following equation:
C₆H₁₂O₆ (s) + 6 O₂ (g) → 6 CO₂ (g) + 6 H₂O (l)
Calculate the volume (in mL) of CO₂ produced at 34°C and 1728.9 torr when 231.88 g glucose is used up in the reaction.

• A. 8.6×10⁵ mL
• B. 6.8×10⁴ mL
• C. 8.6×10⁴ mL
• D. 4.3×10⁴ mL

Answer: (C)

Verified step by step guidance

1. Step 1: Begin by calculating the number of moles of glucose (C6H12O6) used in the reaction. Use the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). Here, the mass of glucose is 231.88 g and the molar mass is 180.156 g/mol.
2. Step 2: Use the stoichiometry of the balanced chemical equation to determine the moles of CO2 produced. According to the equation, 1 mole of glucose produces 6 moles of CO2. Multiply the moles of glucose by 6 to find the moles of CO2.
3. Step 3: Convert the temperature from Celsius to Kelvin using the formula: \( T(K) = T(°C) + 273.15 \). This is necessary for using the ideal gas law.
4. Step 4: Convert the pressure from torr to atm using the conversion factor: \( 1 \text{ atm} = 760 \text{ torr} \). This will allow you to use the ideal gas law.
5. Step 5: Apply the ideal gas law \( PV = nRT \) to find the volume of CO2 produced. Rearrange the formula to solve for volume \( V = \frac{nRT}{P} \). Use \( R = 0.0821 \text{ L atm/mol K} \) as the ideal gas constant. Finally, convert the volume from liters to milliliters by multiplying by 1000.