Combustion vapor-air mixtures are flammable over a limited range of concentrations. The minimum volume % of vapor that gives a combustible mixture is called the lower flammability limit (LFL). Generally, the LFL is about half the stoichiometric mixture, the concentration required to complete combustion of the vapor in air. a) If oxygen is 20.9 vol % of air, estimate the LFL for n-hexane, C6H14. b) What volume of C6H14 (d=.660g/cm^3) is required to produce a flammable mixture of hexane in 1.000m^3 of air STP?

Question

Carolyn Meadows

Chemistry

11 May, 13:17

Combustion vapor-air mixtures are flammable over a limited range of concentrations. The minimum volume % of vapor that gives a combustible mixture is called the lower flammability limit (LFL). Generally, the LFL is about half the stoichiometric mixture, the concentration required to complete combustion of the vapor in air. a) If oxygen is 20.9 vol % of air, estimate the LFL for n-hexane, C6H14. b) What volume of C6H14 (d=.660g/cm^3) is required to produce a flammable mixture of hexane in 1.000m^3 of air STP?

+4

Answers (1)

Know the Answer?

Livia · Answer 1

Combustion equation of n-hexane:

2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O

a)

Assuming we have 100 moles of air,

Oxygen = 20.9 moles

n-hexane required = 20.9/19 x 2

= 2.2 moles

LFL = Half of stoichometric amount = 2.2 / 2 = 1.1

LFL n-hexane = 1.1%

b)

1.1 volume percent required for LFL

1.1% x 1

= 0.0011 m³ of n-hexane required