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The combustion of natural gas is an important process for producing energy. Natural gas primarily consists of methane \((\mathrm{CH}_{4})\) and can also contain other hydrocarbons such as ethane \((\mathrm{C}_{2}\mathrm{H}_{6})\). In this context, the combustion process involves burning these components in the presence of oxygen \((\mathrm{O}_{2})\) to release energy and form carbon dioxide \((\mathrm{CO}_{2})\) and water \((\mathrm{H}_{2}\mathrm{O})\).
Complete combustion of methane requires two molecules of oxygen per molecule of methane:\[\mathrm{CH}_{4} + 2\,\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + 2\,\mathrm{H}_{2}\mathrm{O} \]
For ethane, the reaction is more oxygen-intensive:\[\mathrm{C}_{2}\mathrm{H}_{6} + \frac{7}{2}\,\mathrm{O}_{2} \rightarrow 2\,\mathrm{CO}_{2} + 3\,\mathrm{H}_{2}\mathrm{O} \]
To ensure complete combustion in industrial applications, a calculated amount of excess air is often used. Excess air helps ensure all fuel burns completely and reduces pollutant formation. In our exercise, a 25% excess air is specified to optimize the combustion efficiency. This means taking the total air amount required for stoichiometric combustion and increasing it by 25% to guarantee all the fuel is combusted.

The ideal gas law is a fundamental principle in chemistry and physics that helps us relate the pressure, volume, and temperature of a gas. It's written as:\[ PV = nRT \]
Where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles of gas, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin.
In this exercise, the ideal gas law was used to adjust the flow rate of natural gas from actual conditions to standard conditions, known as standard cubic meters per hour (SCMH). This conversion is necessary because flow meters are typically calibrated to read standard conditions.
To do this, we use the relation:\[ Q_{s} = Q_{a} \times \frac{T_{a}}{T_{s}} \times \frac{P_{s}}{P_{a}} \]
Here, \(T_{a}\) and \(T_{s}\) are actual and standard temperatures, and \(P_{a}\) and \(P_{s}\) are actual and standard pressures respectively. By doing this conversion, we can accurately determine the standard flow rate through the gas flow meter in the system.

Mole fraction is a way to express the concentration of a component in a mixture. It is calculated as the number of moles of a component divided by the total number of moles of all components. The mole fractions in a mixture always sum up to one.
For our natural gas example, the mole fraction of each component is determined using their weight percentages and molar masses. Methane and ethane are the key components: - Molar mass of \(\mathrm{CH}_{4}\) is 16 g/mol - Molar mass of \(\mathrm{C}_{2}\mathrm{H}_{6}\) is 30 g/mol
To calculate the mole fraction of methane \((y_{\mathrm{CH}_{4}})\), we use:\[ y_{\mathrm{CH}_{4}} = \frac{ \frac{95}{16} }{ \frac{95}{16} + \frac{5}{30} } \]
For ethane, the calculation is:\[ y_{\mathrm{C}_{2}\mathrm{H}_{6}} = \frac{ \frac{5}{30} }{ \frac{95}{16} + \frac{5}{30} } \]
These calculations allow us to determine the proportion of each gas in the mixture. A correct mole fraction calculation is crucial for accurate stoichiometric air determination, ensuring efficient and clean combustion.