How many liters of carbon dioxide gas are produced by the complete combustion of 50.0 mol of glucose C₆H₁₂O₆ at 25°C and 1 atm?

Question

| General Chemistry 2

Accepted Answer

Combustion of glucose:

C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O

According to this equation: 1 mole of glucose affords 6 moles of CO₂

⇒ n_CO2 = 6 x n_glucose = 300 mol

Ideal gas law: PV = nRT

with P = pressure (in Pa) à 1 atm = 1.013 x 10⁵ Pa
V = volume (in m³)
n = number of moles (in mol)
R = ideal gas constant = 8.314 (in J.K^-1.mol^-1)
T = temperature (in K) à 0°C = 298 K

V = nRT / P = (300 x 8.314 x 298) / (1.013 x 10⁵)

⇒ V = 7.34 m³ = 7.34 x 10³ dm³ = 7.34 x 10³ L

Midterm 2 - Question 5 | General Chemistry 2

General Chemistry 2 - Midterm 2