Question

Propal alcohol (C \(_{3} \mathrm{H}_{7} \mathrm{OH}\) ) is burned with 50 percent excess air. Write the balanced reaction equation for complete combustion and determine the air-to-fuel ratio.

Short answer

Answer: The air-to-fuel ratio for the combustion of propal alcohol with 50% excess air is 32.14:1.

Step-by-step solution

Write the unbalanced equation for complete combustion of propal alcohol

For complete combustion, propal alcohol will react with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). The unbalanced equation can be written as: C3H7OH + O2 → CO2 + H2O

Balance the chemical equation

To balance the equation, we need to adjust the coefficients so that the number of atoms of each element is equal on both sides of the equation. Starting with carbon: 3 * (C) + O2 → 3 * (CO2) + H2O Next, balance the hydrogen: C3H7OH + O2 → 3 * (CO2) + 4 * (H2O) Finally, balance the oxygen: C3H7OH + 4.5 * (O2) → 3 * (CO2) + 4 * (H2O) Now the equation is balanced: C3H7OH + 4.5O2 → 3CO2 + 4H2O

Account for the excess air

The problem states that there is 50% excess air. Hence, we will multiply the amount of O2 in the balanced equation by 1.5 (100% + 50% = 150% = 1.5): C3H7OH + 6.75O2 → 3CO2 + 4H2O

Calculate the air-to-fuel ratio, considering the stoichiometry of air

Air comprises about 21% oxygen (O2) and 79% nitrogen (N2) by volume. So, for every mole of O2, there will be 79/21 = 3.76 moles of N2. Thus, air required for the combustion of propal alcohol would be: 6.75 * O2 + (6.75 * 3.76) * N2 To find the air-to-fuel ratio, we will calculate the moles of air (moles of O2 and N2) required to burn 1 mole of propal alcohol: 6.75 moles of O2 + 25.39 moles of N2 = 32.14 moles of air The air-to-fuel ratio is then: 32.14 moles of air / 1 mole of propal alcohol

Write the final answer

The balanced reaction equation for the complete combustion of propal alcohol with 50% excess air is: C3H7OH + 6.75O2 → 3CO2 + 4H2O The air-to-fuel ratio is 32.14:1.

Key concepts

Chemical Equation Balancing

Balancing a chemical equation is fundamental in chemistry to adhere to the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a chemical reaction. It means that the atoms of each element must be conserved across the reactants and products of the reaction.

For example, the combustion of propal alcohol is represented by an unbalanced equation: \( C_3H_7OH + O_2 \rightarrow CO_2 + H_2O \). To balance this equation, we adjust the coefficients, ensuring the same number of each type of atom on both sides. We consider each element separately, often starting with the most complex molecule. In the example, we balance carbon (C), hydrogen (H), and finally oxygen (O), which often comes last since it is frequently found in multiple compounds.

Ultimately, this process leads to the balanced equation: \( C_3H_7OH + 4.5O_2 \rightarrow 3CO_2 + 4H_2O \).

Mathematical Considerations in Balancing

Algorithmic methods, like the balancing of algebraic equations, can be helpful when balancing more complex chemical reactions. However, practice and familiarity with the common types of reactions are often the best tools for efficiently balancing chemical equations.

Stoichiometry

Stoichiometry is the area of chemistry that refers to the quantitative relationships between the substances as they participate in chemical reactions. It is based on the balanced chemical equations, such as the one for the combustion of propal alcohol.

In the balanced equation \( C_3H_7OH + 4.5O_2 \rightarrow 3CO_2 + 4H_2O \), stoichiometry tells us that 1 mole of propal alcohol reacts with 4.5 moles of oxygen to produce 3 moles of carbon dioxide and 4 moles of water.

Stoichiometry is crucial for calculating reactant and product quantities, whether in moles, grams, liters, or other units. In advanced applications, stoichiometry can also involve calculations for chemical energy, such as enthalpy or Gibbs free energy, which are related to the thermodynamics of the reaction.

Understanding Mole Ratios

Mole ratios, derived from the coefficients in the balanced chemical equation, provide the foundation for reactant and product relationships. They are used to calculate how much of a reactant is needed or how much of a product is formed.

Air-to-Fuel Ratio

The air-to-fuel ratio (AFR) is a crucial concept in combustion reactions, representing the relationship between the amount of air (containing oxygen) and fuel required to complete a reaction without leaving excess of either. It is particularly important in industrial processes and internal combustion engines for efficiency and pollution control.

In the exercise, we consider an AFR for the complete combustion of propal alcohol with excess air. This involves stoichiometry to determine the amount of oxygen needed and then accounting for the composition of air since air is only 21% oxygen by volume. The balance of air is primarily nitrogen (N2).

When the problem states a 50% excess of air, we calculate the total oxygen based on the stoichiometry of the reaction and then multiply by 1.5 to account for the excess. The presence of nitrogen is significant here as it informs the total volume or mass of air required. Thus, the objective is to determine the volume or mass ratio of air to propal alcohol for the reaction to proceed as stated.

Environmental and Efficiency Impacts

The AFR can significantly impact emissions and combustion efficiency. For instance, too rich (low AFR) or too lean (high AFR) can lead to incomplete combustion, releasing harmful emissions or reducing the energy efficiency of the reaction.