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Combustible 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 flammable limit (LFL). Generally, the LFL is about half the stoichiometric mixture, the concentration required for complete combustion of the vapor in air.
(b) What volume (in mL) of n-hexane (d = 0.660 g/cm³) is required to produce a flammable mixture of hexane in 1.000 m³ of air at STP?

Respuesta :

The volume of n-hexane will be 0.0011 [tex]m^{3}[/tex].

Volume can be determined by using the ideal gas equation. Volume would be a three-dimensional measurement that is employed to gauge a solid shape's capacity. It implies that the volume of a closed form determines how much space this could occupy in three dimensions.

Given data

Combustion equation of n-hexane:

2CH₁₄ + 19O₂ → 12CO₂ + 14H₂O

a)Assuming we have 100 moles of air,

Oxygen = 20.9 moles

n-hexane required =  mass / molar mass =20.9/19 x 2= 2.2 moles

LFL = Half of stoichiometric 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 [tex]m^{3}[/tex].

Therefore, n-hexane required  to produce a flammable mixture of hexane in 0.0011 [tex]m^{3}[/tex].of air at STP.

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