If the lower flammability limit (LFL) for carbon monoxide is 12.5% and for ethane is 2.9%, what is the LFL for a mixture with 0.7 mole fraction of carbon monoxide and 0.3 mole fraction of ethane?

If the lower flammability limit (LFL) for carbon monoxide is 12.5% and for ethane is 2.9%, what is the LFL for a mixture with 0.7 mole fraction of carbon monoxide and 0.3 mole fraction of ethane?

• A. 4.5
• B. 6.27
• C. 8.9
• D. 10.3

Answer: (B)

To determine the lower flammability limit (LFL) for a mixture of two combustible gases, the concept of the flammability limits of individual components must be used in conjunction with their mole fractions. The lower flammability limit of a mixture can be calculated using the formula: \[ LFL_{mix} = (LFL_{1} \times x_{1}) + (LFL_{2} \times x_{2}) \] Where: - \( LFL_{1} \) and \( LFL_{2} \) are the lower flammability limits of the individual gases. - \( x_{1} \) and \( x_{2} \) are the mole fractions of the respective gases in the mixture. For the given problem: - The LFL for carbon monoxide (LFL_CO) is 12.5% or 0.125 as a decimal. - The LFL for ethane (LFL_C2H6) is 2.9% or 0.029 as a decimal. - The mole fraction for carbon monoxide (x_CO) is 0.7, and for ethane (x_C2H6) it is 0.3. Plugging

To determine the lower flammability limit (LFL) for a mixture of two combustible gases, the concept of the flammability limits of individual components must be used in conjunction with their mole fractions. The lower flammability limit of a mixture can be calculated using the formula:

[ LFL_{mix} = (LFL_{1} \times x_{1}) + (LFL_{2} \times x_{2}) ]

Where:

• ( LFL_{1} ) and ( LFL_{2} ) are the lower flammability limits of the individual gases.

• ( x_{1} ) and ( x_{2} ) are the mole fractions of the respective gases in the mixture.

For the given problem:

• The LFL for carbon monoxide (LFL_CO) is 12.5% or 0.125 as a decimal.

• The LFL for ethane (LFL_C2H6) is 2.9% or 0.029 as a decimal.

• The mole fraction for carbon monoxide (x_CO) is 0.7, and for ethane (x_C2H6) it is 0.3.

Plugging