Question

A general formula for gasoline is \(\mathrm{C}_{7} \mathrm{H}_{13}\). Calculate the air to gasoline mass ratio required for stoichiometric combustion.

Short answer

Air to gasoline mass ratio for stoichiometric combustion is approximately 15.32.

Step-by-step solution

Write the Combustion Reaction

The stoichiometric combustion reaction of gasoline, represented by the formula \(\mathrm{C}_{7} \mathrm{H}_{13}\), writes as: \[ \mathrm{C}_{7} \mathrm{H}_{13} + a \mathrm{O}_{2} \rightarrow b \mathrm{CO}_{2} + c \mathrm{H}_{2}\mathrm{O} \] We need to balance this equation to find the coefficient \(a\) in front of \(\mathrm{O}_{2}\), which represents the oxygen from the air.

Balance the Equation

To start balancing, consider the carbon atoms: 7 carbon atoms in \(\mathrm{C}_{7}\mathrm{H}_{13}\) produce 7 \(\mathrm{CO}_{2}\), so \(b = 7\). Then for hydrogen: 13 hydrogen atoms make \(\frac{13}{2}\ \mathrm{H}_{2}\mathrm{O}\), so \(c = \frac{13}{2}\). Now balance the oxygen: On the right side, there are \(7 \times 2 + \frac{13}{2}\) oxygen atoms, so \(a = \frac{14 + \frac{13}{2}}{2} = \frac{19.5}{2} = 9.75\).

Determine the Oxygen Requirement

For the complete combustion of one mole of \(\mathrm{C}_{7} \mathrm{H}_{13}\), \(9.75\) moles of \(\mathrm{O}_{2}\) are required. Each mole of \(\mathrm{O}_{2}\) corresponds to approximately \(32\) grams. Therefore, the mass of oxygen required is \[ 9.75 \times 32 \approx 312\, \text{g} \].

Calculate the Molar Mass of Gasoline

The molar mass of \(\mathrm{C}_{7} \mathrm{H}_{13}\) can be calculated as: \[ 7 \times 12 + 13 \times 1 = 84 + 13 = 97\, \text{g/mol} \]. Thus, the mass of one mole of \(\mathrm{C}_{7} \mathrm{H}_{13}\) is \(97\) g.

Calculate the Air Requirement

Since air is composed of approximately 21% \(\mathrm{O}_{2}\) by volume, to supply \(312\) g of oxygen, the mass of the air needed is \[ \frac{312}{0.21} \approx 1485.71\, \text{g} \].

Calculate the Air to Gasoline Ratio

Finally, the air to gasoline mass ratio for combustion is \[ \frac{1485.71}{97} \approx 15.32 \]. This means about \(15.32\) times more air by mass than gasoline is required for stoichiometric combustion.

Key concepts

Combustion Reaction

A combustion reaction is a chemical process where a substance reacts with oxygen, releasing energy in the form of heat and light. It's a type of exothermic reaction, meaning it gives off energy. In stoichiometric combustion, every atom of the fuel fully reacts with oxygen to produce water and carbon dioxide, without leaving any unburnt fuel or oxygen leftover. This perfect ratio is known as the stoichiometric ratio.
For gasoline, represented by a simplified molecular formula such as \(C_7H_{13}\), the combustion process involves more than just these basic elements. The reaction produces carbon dioxide \(CO_2\) and water \(H_2O\). Understanding the concept of combustion reaction is crucial for applications in engines, where energy needs to be harnessed efficiently.

Chemical Equation Balancing

Balancing a chemical equation is like ensuring both sides of a scale are equal. In chemistry, this means having the same number of each type of atom on both sides of the equation. This task is essential for understanding how much of each reactant is needed and how much product is produced.
To balance the combustion equation for gasoline (\(C_7H_{13}\)), you follow these steps:

• Identify the number of carbon, hydrogen, and oxygen atoms in the reactants and products.
• Adjust coefficients in front of formulas to ensure the atom count for each element is the same on both sides of the equation.
• For carbon and hydrogen atoms first: Carbon atoms in the fuel create \(CO_2\) molecules, while hydrogen atoms create \(H_2O\) molecules.
• Finally, balance the oxygen atoms.

Balancing ensures the law of conservation of mass is obeyed, which states mass is neither created nor destroyed in a chemical reaction.

Gasoline Molecular Formula

The molecular formula of gasoline can vary, but a common representation is \(C_7H_{13}\). This formula tells us there are 7 carbon atoms and 13 hydrogen atoms in a single molecule of this gasoline model.
While actual gasoline is a mixture of many hydrocarbons, this simple formula allows us to approximate its combustion behavior. Understanding the molecular formula is crucial for calculating the stoichiometric combustion ratios, including the amount of air required for complete combustion.
The molar mass, calculated by adding the atomic masses of carbon and hydrogen, is 97 g/mol. This value is vital for converting between moles and grams, which is necessary when working with balanced chemical equations and for finding the Air to Fuel Ratio.

Air to Fuel Ratio

The Air to Fuel Ratio (AFR) is crucial in combustion reactions, especially in engines. It describes the amount of air in relation to the amount of fuel used in combustion. The ratio is expressed in terms of mass.
An ideal or stoichiometric AFR ensures complete combustion without any excess air or fuel left over. In our case, the AFR for gasoline \(C_7H_{13}\) is approximately 15.32. This means for every gram of fuel, 15.32 grams of air are required for complete combustion.
Maintaining the correct AFR is important for optimizing engine performance, reducing emissions, and improving fuel efficiency. A stoichiometric AFR ensures that no fuel is wasted, leading to cleaner combustions and better energy output. It's crucial for meeting environmental standards and maximizing the lifespan of combustion engines.