Question

Propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) is burned with 150 percent theoretical air. The air-fuel mass ratio for this combustion process is \((a) 5.3\) \((b) 10.5\) \((c) 15.7\) \((d) 23.4\) \((e) 39.3\)

Short answer

Answer: The air-fuel mass ratio is approximately 5.3.

Step-by-step solution

Write down the balanced chemical equation for complete combustion of propane.

The balanced chemical equation for complete combustion of propane is: C3H8 + 5O2 → 3CO2 + 4H2O

Calculate the number of moles for the excess air condition.

Since we have 150% theoretical air, we need to multiply the moles of oxygen in the balanced equation by 1.5: C3H8 + 1.5 * 5O2 → 3CO2 + 4H2O This simplifies to: C3H8 + 7.5O2 → 3CO2 + 4H2O

Compute the masses of fuel and air in the combustion process.

Now, let's calculate the masses involved in the combustion process. Molar mass of propane (C3H8): 3(12.01) + 8(1.01) = 44.11 g/mol Molar mass of oxygen (O2): 2(16) = 32 g/mol Using the stoichiometry from the chemical equation, for every 1 mole of propane burned, 7.5 moles of oxygen are required: Mass of propane = 44.11 g/mol Mass of oxygen = 7.5 * 32 g/mol = 240 g/mol

Calculate the air-fuel mass ratio.

The air-fuel mass ratio is the mass of air divided by the mass of fuel: Air-fuel mass ratio = Mass of oxygen / Mass of propane Air-fuel mass ratio = 240 g/mol / 44.11 g/mol = 5.44 Since 5.44 is closest to option (a) 5.3, the air-fuel mass ratio for this combustion process is approximately 5.3.

Key concepts

Stoichiometry

Stoichiometry is a foundational concept in chemistry that provides a systematic way to predict the amounts of substances consumed and produced in chemical reactions. It is based on the principle that matter cannot be created or destroyed in a chemical reaction, which is known as the law of conservation of mass.

At the heart of stoichiometry are balanced chemical equations, which indicate the ratios of reactants and products participating in the reaction. For instance, when looking at the combustion of propane \( \mathrm{C}_{3}\mathrm{H}_{8} \), stoichiometry allows us to determine the precise amount of oxygen needed for complete combustion and the volume of the resulting carbon dioxide and water.

Practical uses of stoichiometry include calculating the fuel efficiency in combustion engines and determining the optimal reactant proportions in industrial chemical manufacturing. To solve stoichiometry problems, one typically follows a methodical approach: identifying the balanced chemical equation, determining the molar relationships between reactants and products, and finally calculating the mass or volume of the substances involved.

Theoretical Air Combustion

Theoretical air refers to the ideal quantity of air needed for the complete combustion of a specific amount of fuel without any excess. In theoretical air combustion, the air supply is precisely that needed to fully react with the fuel, based on the stoichiometry of the chemical reaction. For example, in the combustion of propane, the stoichiometry tells us that one mole of propane requires five moles of oxygen to produce carbon dioxide and water.

In reality, complete combustion is rarely achieved with theoretical air alone due to imperfect mixing and other practical limitations. As a result, combustion typically takes place with some excess air to ensure complete burning of the fuel. In the given problem, 150 percent of theoretical air is used, which means that there is 50 percent more air than what the stoichiometry requires for complete combustion. The presence of excess air ensures that all the fuel has been consumed, while the amount of excess indicates how much the actual air supply exceeds the theoretical minimum required.

Air-Fuel Mass Ratio

The air-fuel mass ratio is a crucial parameter in combustion processes, particularly within engines and industrial burners. It represents the mass of air divided by the mass of fuel in a combustion reaction. This ratio is vital because it influences the temperature of the combustion, the efficiency of fuel usage, and the emissions produced.

To calculate this ratio, one would use the masses of the air (which in combustion is largely made up of oxygen) and fuel as determined through stoichiometry. An optimal air-fuel mass ratio allows for complete and efficient fuel combustion, while deviations from this can result in unburnt fuel (running rich) or excess oxygen (running lean), leading to various performance and environmental impacts. In the exercise solution provided, this ratio was calculated based on 150 percent theoretical air, indicating an excess, which is often the case in real-world applications to ensure complete combustion.