The flame produced by the burner of a gas (propane) grill is a pale blue color when enough air mixes with the propane (C3H8) to burn it completely. For every gram of propane that flows through the burner, what volume of air is needed to burn it completely? Assume that the temperature of the burner is 195.0 Celcius, the pressure is 1.1 atmosphere, and the mole fraction of O2 in air is 0.21.

At the burner temp. and pressure, 18.85 litres of air is needed to completely combust each gram of propane

Explanation:

The combustion stoichiometry is as follows:

C₃H₈ + 5O₂ = 4 H₂O + 3CO₂ The molecular weights (g/mol) are:

MW 44 5x32 4x18 3x44

So each gram of propane is 1/44 = 0.02272 mol propane

and will need 5 x 0.02272 = 0.1136 mol oxygen

At 0.21 mol fraction oxygen in air, 0.1136 / 0.21 = 0.54 mol air is needed to burn the propane.

At the low pressure in the burner we can use the Ideal Gas Law

PV=nRT, or V = nRT/P

P = 1.1 x 101325 Pa = 111457 Pa

T = 195°C + 273 = 468 K

R = 8.314

and we calculated n = number of moles air = 0.54 mol

So V m³ = 0.54 x 8.314 x 468 / 111457 = 0.0188 m³ = 18.85 litres air.