Problem 106 - Chemistry The Central Science In Si Units

Question

Assume that a single cylinder of an automobile engine has a volume of \(524 \mathrm{~cm}^{3}\). (a) If the cylinder is full of air at \(74^{\circ} \mathrm{C}\) and \(99.3 \mathrm{kPa}\), how many moles of \(\mathrm{O}_{2}\) are present? (The (b) How many mole fraction of \(\mathrm{O}_{2}\) in dry air is \(0.2095 .\) ) grams of \(\mathrm{C}_{8} \mathrm{H}_{18}\) could be combusted by this quantity of \(\mathrm{O}_{2}\), assuming complete combustion with formation of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} ?\)

Step-by-Step Solution

Answer

The engine cylinder contains about 0.00377 moles of \( \mathrm{O}_{2} \), capable of combusting approximately 0.0343 grams of octane.

Step 1: Convert Temperature to Kelvin

The given temperature is \( 74^{\circ} \mathrm{C} \). To convert it to Kelvin, use the formula: \( T(K) = T(^{\circ} \mathrm{C}) + 273.15 \). Therefore, the temperature in Kelvin is \( 74 + 273.15 = 347.15 \mathrm{~K} \).

Step 2: Use Ideal Gas Law to Find Moles of Air

The volume of the cylinder is \( V = 524 \mathrm{~cm}^{3} = 0.524 \mathrm{~L} \). The Ideal Gas Law \( PV = nRT \) can be rearranged to solve for \( n \), the number of moles: \( n = \frac{PV}{RT} \). The pressure \( P \) is \( 99.3 \mathrm{kPa} = 99.3 \times 10^3 \mathrm{~Pa} \), and \( R = 8.314 \mathrm{~J/mol} \cdot \mathrm{K} \) (with \( 1 \mathrm{~Pa} \cdot \mathrm{m}^3 = 1 \mathrm{~J} \)). Therefore, \( n = \frac{99.3 \times 10^3 \times 0.000524}{8.314 \times 347.15} \approx 0.018 \) moles of air.

Step 3: Calculate Moles of Oxygen

Given the mole fraction of \( \mathrm{O}_{2} \) in dry air is \( 0.2095 \), the number of moles of \( \mathrm{O}_{2} \) is \( n_{\mathrm{O}_2} = 0.2095 \times 0.018 \approx 0.00377 \) moles.

Step 4: Write and Balance the Combustion Reaction

The complete combustion of \( \mathrm{C}_8 \mathrm{H}_{18} \) is described by the reaction: \( 2 \mathrm{C}_8 \mathrm{H}_{18} + 25 \mathrm{O}_2 \rightarrow 16 \mathrm{CO}_2 + 18 \mathrm{H}_2 \mathrm{O} \). This shows that 2 moles of \( \mathrm{C}_8 \mathrm{H}_{18} \) react with 25 moles of \( \mathrm{O}_2 \).

Step 5: Determine Moles of \( \mathrm{C}_8 \mathrm{H}_{18} \) that can be Combusted

From the balanced equation, 1 mole of \( \mathrm{C}_8 \mathrm{H}_{18} \) requires \( 12.5 \) moles of \( \mathrm{O}_2 \). Therefore, \( 0.00377 \) moles of \( \mathrm{O}_2 \) can combust \( \frac{0.00377}{12.5} \approx 0.00030 \) moles of \( \mathrm{C}_8 \mathrm{H}_{18} \).

Step 6: Convert to Grams of \( \mathrm{C}_8 \mathrm{H}_{18} \)

The molar mass of \( \mathrm{C}_8 \mathrm{H}_{18} \) (octane) is \( 114.22 \mathrm{~g/mol} \). Therefore, the mass of octane that can be combusted is \( 0.00030 \times 114.22 \approx 0.0343 \mathrm{~g} \).

Key Concepts

Stoichiometry

Stoichiometry is a key concept in chemistry that deals with the calculation of reactants and products in chemical reactions. It involves using balanced chemical equations to determine the relationships between different chemicals involved in a reaction. In simple terms, stoichiometry allows us to predict how much of each substance is consumed or produced during a reaction.

In this exercise, we deal with the combustion of octane, \(\mathrm{C}_8 \mathrm{H}_{18}\). The balanced equation for this reaction is:\[2 \mathrm{C}_8 \mathrm{H}_{18} + 25 \mathrm{O}_2 \rightarrow 16 \mathrm{CO}_2 + 18 \mathrm{H}_2 \mathrm{O}\]This tells us that 2 moles of octane react with 25 moles of oxygen to produce 16 moles of carbon dioxide and 18 moles of water.

By knowing the balanced equation, we can relate the amount of oxygen available to the amount of octane that can be combusted. Stoichiometry is an essential tool to calculate these relationships to ensure reactions are carried out efficiently, without excess reactants.

Mole Fraction

The mole fraction of a component in a mixture indicates how much of that component is present in terms of moles, compared to the total number of moles of all components. It is a dimensionless quantity used to express the concentration of a component in a mixture.

The mole fraction is calculated as:\[\text{Mole Fraction of } \mathrm{O}_2 = \frac{\text{Moles of } \mathrm{O}_2}{\text{Total Moles of Air}}\]In this exercise, the mole fraction of \(\mathrm{O}_2\) in air is given as \(0.2095\). This means that approximately 20.95% of the air's moles consist of oxygen. By knowing the total moles of air, we can calculate the moles of oxygen using the mole fraction.

This is a crucial concept in various fields such as atmospheric science, chemical engineering, and combustion analysis where knowing the proportion of a specific gas in a mixture is essential.

Combustion Reaction

A combustion reaction is a chemical process where a substance reacts with oxygen to produce heat and light. Typically, hydrocarbons like octane (\(\mathrm{C}_8 \mathrm{H}_{18}\)) are common fuels that undergo combustion. These reactions are vital for generating energy in engines, power stations, and more.

The combustion of octane can be represented by the reaction:\[2 \mathrm{C}_8 \mathrm{H}_{18} + 25 \mathrm{O}_2 \rightarrow 16 \mathrm{CO}_2 + 18 \mathrm{H}_2 \mathrm{O}\]In this reaction, octane burns in the presence of oxygen producing carbon dioxide and water as the main by-products. This is an exothermic reaction, meaning it releases energy in the form of heat.

Understanding combustion reactions is crucial for designing engines and fuels that operate efficiently and minimize environmental impact. Overall, combustion reactions convert chemical energy into mechanical or thermal energy, driving many of the world's technological processes.