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. First, calculate the number of moles of glucose (C6H12O6) used in the reaction. Use the formula: \( \text{moles of glucose} = \frac{\text{mass of glucose}}{\text{molar mass of glucose}} \). Given that the mass of glucose is 231.88 g and the molar mass is 180.156 g/mol, substitute these values into the formula.
2. According to the balanced chemical equation, 1 mole of glucose produces 6 moles of CO2. Use the stoichiometry of the reaction to calculate the moles of CO2 produced: \( \text{moles of CO2} = 6 \times \text{moles of glucose} \).
3. Next, use the ideal gas law to find the volume of CO2 produced. The ideal gas law is \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is moles, \( R \) is the ideal gas constant, and \( T \) is temperature in Kelvin. Convert the temperature from Celsius to Kelvin by adding 273.15 to the Celsius temperature.
4. Convert the pressure from torr to atm, since the ideal gas constant \( R \) is typically given in L·atm/(mol·K). Use the conversion: \( 1 \text{ atm} = 760 \text{ torr} \).
5. Rearrange the ideal gas law to solve for volume \( V \): \( V = \frac{nRT}{P} \). Substitute the values for \( n \) (moles of CO2), \( R \) (0.0821 L·atm/(mol·K)), \( T \) (temperature in Kelvin), and \( P \) (pressure in atm) into the equation to calculate the volume in liters. Finally, convert the volume from liters to milliliters by multiplying by 1000.