Question

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, which is the concentration required for complete combustion of the vapor in air. (a) If oxygen is 20.9 vol \% of air, estimate the LFL for \(n\) -hexane, \(\mathrm{C}_{6} \mathrm{H}_{14}\). (b) What volume (in \(\mathrm{mL}\) ) of \(n\) -hexane \(\left(d=0.660 \mathrm{~g} / \mathrm{cm}^{3}\right)\) is required to produce a flammable mixture of hexane in \(1.000 \mathrm{~m}^{3}\) of air at \(\mathrm{STP} ?\)

Short answer

a) LFL for hexane is 1.1 vol %. b) 64.1 mL of n-hexane is required.

Step-by-step solution

- Write the Combustion Reaction

The combustion reaction for hexane (\text{C}_{6}\text{H}_{14}) is: \[\text{C}_{6}\text{H}_{14} + \text{O}_{2} \rightarrow \text{CO}_{2} + \text{H}_{2}\text{O}\] Balance the equation to ensure you have the correct stoichiometric mixture.

- Balance the Combustion Reaction

Balance the combustion reaction: \[ \text{C}_{6}\text{H}_{14} + \frac{19}{2}\text{O}_{2} \rightarrow 6\text{CO}_{2} + 7\text{H}_{2}\text{O} \].

- Calculate Stoichiometric Volume

Using the balanced equation, determine the stoichiometric volume of hexane. Since oxygen is 20.9\text{\text{vol %}} of air, the stoichiometric volume of hexane is: \[ \frac{20.9\text{\text{vol %}}}{2 \times 19 / 2} = 2.2\text{vol %} \]

- Estimate the Lower Flammable Limit

The LFL is generally about half the stoichiometric volume. Therefore, LFL for hexane is: \[LFL = \frac{2.2\text{\text{vol %}}}{2} = 1.1\text{\text{vol %}} \]

- Convert Air Volume to moles using STP

Given 1.000 \text{m}^{3} of air at STP, converting to moles using the ideal gas law: \[ 1.000 \text{m}^{3} = 1,000 \text{L} \] \[ \text{at STP: } 1 \text{mol} = 22.4 \text{L} \rightarrow \frac{1,000 L}{22.4 L/mol} \rightarrow 44.64 \text{moles }\]

- Calculate Volume of n-hexane for Flammable Mixture

Determine the mole fraction of LFL concentration: \[ 1.1\text{\text{vol %}} \rightarrow 0.011 \times 44.64 \text{ moles } = 0.491\text{ moles } \] Now convert moles of hexane to volume (using density and molar mass): \[ \text{Molar mass of C}_{6}\text{H}_{14} = 86.18 \text{g/mol} \] \[ \text{Volume =}\frac{\text{Mass}}{\text{Density}}:\] \[ \text{Mass }= 0.491\text{ moles} \times 86.18 \text{g/mol} = 42.31 \text{g } \] \[ \text{Volume } = \frac{42.31g}{0.660g/cm^3} \] \[ \text{Volume }= 64.1 \text{mL}\]

Key concepts

Combustion Reaction

Combustion reactions are fundamental chemical processes where a fuel reacts with an oxidant, releasing energy in the form of heat and light. For n-hexane (\text{C}_{6}\text{H}_{14}), the typical combustion reaction in the presence of oxygen (\text{O}_{2}) can be represented as:
\[\text{C}_{6}\text{H}_{14} + \text{O}_{2} \rightarrow \text{CO}_{2} + \text{H}_{2}\text{O}\]
This represents the initial unbalanced form of the equation. Balancing the equation is crucial to understanding the stoichiometric relationships—ratios that allow all reactants to completely react without any excess.
After balancing, we get:
\[\text{C}_{6}\text{H}_{14} + \frac{19}{2}\text{O}_{2} \rightarrow 6\text{CO}_{2} + 7\text{H}_{2}\text{O}\]
In this balanced reaction, 1 mole of hexane requires 9.5 moles of oxygen to fully combust, producing 6 moles of carbon dioxide (\text{CO}_{2}) and 7 moles of water (\text{H}_{2}\text{O}). This balanced equation is essential for calculating flammability limits and volumes.

Stoichiometric Mixture

A stoichiometric mixture refers to the perfect ratio of fuel and oxidant that results in complete combustion, with no leftover reactants. For n-hexane in air, this ratio is key to determining the lower flammable limit (LFL). With oxygen comprising 20.9% of the air, we calculate the stoichiometric volume of hexane:
\[\frac{20.9 \text{vol %}}{2 \times 9.5} = 2.2 \text{vol %}\]
This means that in a perfect mixture, 2.2% of the volume would be hexane, and the rest would be air, ensuring complete combustion. The LFL is generally about half of this stoichiometric volume to ensure a safe margin against incomplete combustion. Therefore, \[\text{LFL} = \frac{2.2 \text{vol %}}{2} = 1.1 \text{vol %}\]
Thus, 1.1% volume of hexane in air is the minimum required for a flammable mixture, ensuring ignition and sustained combustion when sparked.

Ideal Gas Law

The Ideal Gas Law is instrumental in converting volumes and masses into moles. It's given by: \[\text{PV = nRT}\] where P = pressure, V = volume, n = number of moles, R = gas constant, and T = temperature. Under standard temperature and pressure (STP), this simplifies to:
\[\text{1 mole of gas occupies 22.4 L}\]
Given 1.000 m\text{3} of air at STP:
\[\text{1.000 m}^{3} = 1,000 \text{L}\]
we can convert this volume to moles:
\[\frac{1,000 \text{L}}{22.4 \text{L/mol}} = 44.64 \text{moles}\]
This value represents the number of moles of air. By considering hexane's LFL (1.1 vol %), we calculate the moles of hexane needed:
\[\text{0.011} \times 44.64 = 0.491 \text{moles}\]
This conversion aids in further calculations, especially in determining mass and volume of fuel required for combustion.

Density and Molar Mass Conversion

To determine the volume of hexane needed for a flammable mixture, we convert moles to mass using its molar mass, and then mass to volume using density. The molar mass of n-hexane (\text{C}_{6}\text{H}_{14}) is:
\[\text{Molar mass} = 86.18 \text{g/mol}\]
Using the calculated moles (0.491), the mass is:
\[\text{Mass} = 0.491 \text{moles} \times 86.18 \text{g/mol} = 42.31 \text{g}\]
Now, to convert mass to volume, we use the hexane density (0.660 g/cm³):
\[\text{Volume} = \frac{42.31 \text{g}}{0.660 \text{g/cm}^3} = 64.1 \text{mL}\]
Therefore, 64.1 mL of n-hexane is needed in 1.000 m³ of air to produce a flammable mixture at the lower flammable limit. This conversion process ensures accurate and practical applications in real-world situations involving flammable substances.