Question

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?

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

Answer 2

(a) The LFL for n-hexane is   $\boxed{1.1\;\% }$

(b) $\boxed{{\text{0}}{\text{.011 }}{{\text{m}}^{\text{3}}}}$ of n-hexane is required to produce its own flammable mixture in ${\text{1 }}{{\text{m}}^{\text{3}}}$  of air.

Further explanation:

Stoichiometry of a reaction is used to determine the amount of species present in reaction by the relationship between reactants and products. It is used to determine moles of a chemical species when moles of other chemical species present in reaction is given.

Consider the general reaction,

${\text{A}}+2{\text{B}}\to3{\text{C}}$

Here,

A and B are reactants.

C is the product.

One mole of A reacts with two moles of B to produce three moles of C. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3, and the stoichiometric ratio between B and C is 2:3.

Combustion reactions:

These are the reactions that take place when hydrocarbons are burnt in the presence of oxygen to form carbon dioxide and water. These are also referred to as burning.

Example of combustion reactions are as follows:

(a)   ${\text{C}}{{\text{H}}_4}+{{\text{O}}_2}\to{\text{C}}{{\text{O}}_2}+{{\text{H}}_2}{\text{O}}$

(b) ${{\text{C}}_{10}}{{\text{H}}_{14}}+12{{\text{O}}_2}\to10{\text{C}}{{\text{O}}_2}+4{{\text{H}}_2}{\text{O}}$

Lower Flammability Limit:

It is the minimum volume % of vapor that produces a combustible mixture. It is usually half the stoichiometric mixture concentration that is required for complete combustion of vapor in the air. It is written as LFL.

(a) The balanced chemical equation for combustion of n-hexane is as follows:

${\text{2}}{{\text{C}}_6}{{\text{H}}_{14}}+19{{\text{O}}_2}\to12{\text{C}}{{\text{O}}_2}+14{{\text{H}}_{\text{2}}}{\text{O}}$

From the balanced chemical reaction, the reaction stoichiometry between ${{\text{C}}_6}{{\text{H}}_{14}}$  and ${{\text{O}}_2}$ is as follows:

$2{\text{ mol }}{{\text{C}}_6}{{\text{H}}_{14}}:19{\text{ mol }}{{\text{O}}_2}$

Consider 100 moles of air. The moles of oxygen can be calculated as follows;

${\text{Moles of }}{{\text{O}}_{\text{2}}}=\left({\frac{{{\text{Volume\%  of }}{{\text{O}}_{\text{2}}}}}{{{\text{100\%}}}}}\right)\left({{\text{Moles of air}}}\right)$                   …… (1)

The volume % of ${{\text{O}}_2}$  is 20.9 %.

The moles of air are 100 mol.

Substitute these values in equation (1).

$\begin{aligned}{\text{Moles of }}{{\text{O}}_{\text{2}}}&=\left({\frac{{{\text{20}}{\text{.9\%}}}}{{{\text{100 \% }}}}}\right)\left({{\text{100 mol}}}\right)\\&={\text{20}}{\text{.9 mol}}\\\end{aligned}$

19 moles of ${{\text{O}}_2}$  require 2 moles of n-hexane. So the amount of n-hexane required for 20.9 moles of ${{\text{O}}_2}$  is calculated as follows:

$\begin{aligned}{\text{Moles of }}{{\text{C}}_6}{{\text{H}}_{14}}&=\left( {{\text{20}}{\text{.9 mol }}{{\text{O}}_2}}\right)\left({\frac{{{\text{2 mol }}{{\text{C}}_6}{{\text{H}}_{14}}}}{{{\text{19}}\;{\text{mol }}{{\text{O}}_{\text{2}}}}}}\right)\\&={\text{2}}{\text{.2 mol}}\\\end{aligned}$

The formula to calculate LFL for n-hexane is calculated as follows:

${\text{LFL for }}{{\text{C}}_6}{{\text{H}}_{14}}=\frac{{{\text{Stoichiometric amount of }}{{\text{C}}_6}{{\text{H}}_{14}}}}{2}$                   …… (2)

Substitute 2.2 mol for the stoichiometric amount of ${{\text{C}}_6}{{\text{H}}_{14}}$  in equation (2).

$\begin{aligned}{\text{LFL for }}{{\text{C}}_6}{{\text{H}}_{14}}&=\frac{{{\text{2}}{\text{.2 mol}}}}{2}\\&={\text{1}}{\text{.1 mol}}\\\end{aligned}$

So LFL for n-hexane is 1.1 %.

(b) The LFL value for ${{\text{C}}_6}{{\text{H}}_{14}}$  calculated in part (a) is used to find its volume. This is done by using equation (3).

The formula to calculate the volume of ${{\text{C}}_6}{{\text{H}}_{14}}$  is as follows:

${\text{Volume of }}{{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{14}}}}=\left( {\frac{{{\text{LFL}}\left( {{\text{vol \% }}}\right){\text{for }}{{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{14}}}}}}{{100\;\% }}}\right)\left({{\text{Volume of air}}}\right)$            

…… (3)

The LFL for ${{\text{C}}_6}{{\text{H}}_{14}}$  is 1.1 %.

The volume of air is ${\text{1 }}{{\text{m}}^{\text{3}}}$ .

Substitute these values in equation (3).

$\begin{aligned}{\text{Volume of }}{{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{14}}}}&=\left({\frac{{{\text{1}}{\text{.1 \% }}}}{{100\;\%}}}\right)\left({{\text{1}{{\text{m}}^3}}\right)\\&={\text{0}}{\text{.011}}\;{{\text{m}}^{\text{3}}}\\\end{aligned}$

Learn more:

1. Calculate the moles of chlorine in 8 moles of carbon tetrachloride: https://brainly.com/question/3064603

2. Calculate the moles of ions in the solution: https://brainly.com/question/5950133

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Mole concept

Keywords: stoichiometry, reactant, product, combustion, n-hexane, H2O, CO2, C6H14, LFL, stoichiometric amount, 1.1 %, 20.9 %, 20.9 mol, air, O2.