Question

How many pounds of air (which is \(23.19 \% \mathrm{O}_{2}\) and \(75.46 \%\) \(\mathrm{N}_{2}\) by weight) would be needed to burn a pound of gasoline by a reaction whereby \(\mathrm{C}_{8} \mathrm{H}_{18}\) reacts with \(\mathrm{O}_{2}\) to form \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) ?

Short answer

About 6.86 kg (15.1 pounds approx) of air are needed to burn one pound of gasoline in a combustion reaction that produces CO2 and H2O.

Step-by-step solution

Write the balanced chemical equation for the combustion reaction

Consider the combustion of octane (C8H18) with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). The unbalanced chemical equation is: \( C_8H_{18} + O_2 \rightarrow CO_2 + H_2O \) To balance this equation, we need 8 moles of CO2 on the product side (for the 8 carbon atoms in C8H18) and 9 moles of H2O (for the 18 hydrogen atoms in C8H18). Since there are 25 oxygen atoms on the product side (16 from CO2 and 9 from H2O), we need 12.5 moles of O2 on the reactant side. Therefore, the final balanced chemical equation is: \( C_8H_{18} + 12.5 O_2 \rightarrow 8 CO_2 + 9 H_2O \)

Determine the stoichiometry of the reaction

In the balanced chemical equation, we have: 1 mole of C8H18 reacts with 12.5 moles of O2 to produce 8 moles of CO2 and 9 moles of H2O.

Calculate the amount of O2 required to burn one pound of C8H18

First, convert the mass of C8H18 to moles using its molecular weight: Molecular weight of C8H18: 8 × 12.01 (carbon) + 18 × 1.01 (hydrogen) = 114.23 g/mol Since we have 1 pound of C8H18, convert that to grams: 1 pound = 453.59 grams Moles of C8H18 = (453.59 g) / (114.23 g/mol) = 3.97 mol (approx) Now, to find out the moles of O2 needed based on the stoichiometry, we multiply the moles of C8H18 by the ratio of O2 moles to C8H18 moles: Moles of O2 = 3.97 mol C8H18 × (12.5 mol O2 / 1 mol C8H18) = 49.63 mol O2 (approx) Convert the moles of O2 to grams using the molecular weight of O2 (32.00 g/mol): Mass of O2 = (49.63 mol) × (32.00 g/mol) = 1588.16 g = 1.59 kg (approx)

Translate this amount of O2 to weight of air needed based on air composition

The composition of air states that 23.19% is O2 by weight. Therefore, we can calculate the weight of air needed by using the formula: Weight of air = (Weight of O2 needed) / (Percentage of O2 in air) Weight of air = (1.59 kg) / (23.19/100) = 6.86 kg (approx) So, about 6.86 kg (15.1 pounds approx) of air are needed to burn one pound of gasoline.

Key concepts

Chemical Equation Balancing

Understanding the balancing of chemical equations is essential when studying combustion reactions. A chemical equation represents a reaction where the number of atoms for each element must be the same on both sides. For the combustion of octane, denoted by the formula C₈H₁₈, it's vital to determine the appropriate amount of oxygen (O₂) needed for the reaction.

To balance this equation, we acknowledge the conservation of mass principle, which states that atoms are neither created nor destroyed in chemical reactions. This entails arranging coefficients - simple whole numbers placed in front of the compounds - to ensure that the number of atoms for each element is the same on the reactants and products side. For octane combustion, this involves confirming that there are 8 carbon (C) and 18 hydrogen (H) atoms, requiring 8 molecules of carbon dioxide (CO₂) and 9 of water (H₂O) to preserve the balance. With these substances balanced, we can then calculate the necessary moles of oxygen O₂, considering it will combine with carbon and hydrogen to form the products.

Mole to Mass Conversion

Moving from the abstract concept of moles to the tangible concept of mass is a critical step in stoichiometry. A mole is a unit that chemists use to express amounts of a chemical substance. To convert moles to mass, one must use the molecular weight of the substance, which is the sum of the atomic weights of all the atoms in the molecule expressed in grams per mole (g/mol).

Given that the molecular weight of octane (C₈H₁₈) is 114.23 g/mol, you can calculate the number of moles in one pound of octane by converting the mass from pounds to grams and dividing it by the molecular weight. This step elucidates how much of a substance is available for a reaction in quantitative terms, which can then be used to figure out the mass to complete the reaction - in this case, how much oxygen is needed in mass to burn the given amount of octane.

Combustion of Hydrocarbons

Hydrocarbons like octane are compounds made of hydrogen and carbon, and combustion is their chemical reaction with oxygen that releases energy. When hydrocarbons burn in the presence of oxygen, the result is primarily carbon dioxide (CO₂) and water (H₂O).

This exothermic process (which releases heat) is the underlying principle of engines and power stations that use hydrocarbon-based fuels. To calculate the air required for the complete combustion of a hydrocarbon, it is important to ensure that we supply enough oxygen for all carbon and hydrogen atoms within the fuel to oxidize. Incomplete combustion, which can occur with insufficient oxygen, produces carbon monoxide (CO) and can be harmful. Therefore, accurate calculations achieved through stoichiometry assure the efficiency and safety of hydrocarbon combustion processes.

Percentage Composition by Weight

The percentage composition by weight of a compound or mixture is a way to express the proportion of each element or component by mass. This is particularly relevant when dealing with mixtures like air, which is composed of different gases, each contributing to the total mass based on its percentage.

In our exercise involving the combustion of gasoline, air is composed of 23.19% oxygen and 75.46% nitrogen by weight. If we know the mass of oxygen required for combustion, we can calculate the total weight of air needed by considering the percentage composition of oxygen in air. This step ensures that a sufficient amount of air is provided for combustion, taking into account that air is not pure oxygen and that the remaining major component, nitrogen, does not react with the hydrocarbons during the combustion process. Establishing the percentage composition by weight facilitates the determination of each component's contribution to a reaction or process.