Question

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 (L) of air is needed to burn it completely? Assume that the temperature of the burner is 195.0°C, the pressure is 1.05 atm, and the mole fraction of O₂ in air is 0.210.

Answer 1

writing the reaction; C3 Hg + 50g – 3 CO₂ + 4H₂O For 19 propane, we have - 19C₃H8 x 1 mol C₂ H₂ 44.09629 = 0.0227 mols Cz He Mol ratio of C₂ H₂ to oa is 1:5 So 5 x 0.0227 = 0.1135 mols of Oz 28. needed to react completely with this amount of propane. Guven Mole fraction of no 0.210 ( total = 1)

Here, - 0. 1135 molsa, 0.210 7 ?? mols aire on calculation; we get [0.54 mols of aer, which is needed mole fraction. for this Finally, we calculate volume of als in Lutre PV = nRT Here D=1.05 atm; n = 0.54 mols aire R un Litre = 0.0821 L. atm/mork Temperature in k = 195+273 = 468k (0.54 mots air) (0.821 L. ateismolk) (468k) 1.05 atm [V = 19.76 L of air
here, we have first written the equation.

then found the value of n.

as we know the formula PV=nRT

from there we got V=nRT/P

substituting the values we get volume of air=19.76 L