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 \(0.980 \mathrm{~atm}\), how many moles of \(\mathrm{O}_{2}\) are present? (The mole fraction of \(\mathrm{O}_{2}\) in dry air is \(0.2095 .\) ) (b) How many 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 \(\mathrm{H}_{2} \mathrm{O}\) ?

Short answer

(a) There are \(4.60 \times 10^{-3}\) moles of O₂ present in the cylinder. (b) 0.042 grams of C₈H₁₈ could be combusted by this quantity of O₂, assuming complete combustion.

Step-by-step solution

Convert given temperature to Kelvin

First, convert the given temperature from Celsius to Kelvin by adding 273.15. This will be useful in the ideal gas law calculations. \(T(K) = T(°C) + 273.15\) \(T(K) = 74°C + 273.15\) \(T(K) = 347.15 K\)

Calculate the moles of air in the cylinder

Next, apply the ideal gas law formula (PV = nRT) to determine the number of moles of air. We'll need to rearrange the formula to solve for moles (n): n = PV/RT We're given: P = 0.980 atm V = 524 cm³ (convert to Liters: 524 cm³ × (1 L / 1000 cm³) = 0.524 L) R = 0.0821 L atm/K mol (ideal gas constant) T = 347.15 K (from Step 1) Plug the values into the formula: n (air) = (0.980 atm × 0.524 L) / (0.0821 L atm/K mol × 347.15 K) n (air) = 0.0219 mol

Calculate the moles of O₂ in the cylinder

Use the mole fraction of O₂ in dry air (0.2095) to determine the moles of O₂ in the cylinder: n (O₂) = mole fraction × n (air) n (O₂) = 0.2095 × 0.0219 mol n (O₂) = 4.60 × 10⁻³ mol

Obtain balanced combustion reaction

Write down the balanced reaction for the complete combustion of C₈H₁₈: C₈H₁₈ + 12.5 O₂ → 8 CO₂ + 9 H₂O

Calculate the moles and grams of C₈H₁₈ combusted

Use stoichiometry and the balanced equation in Step 4 to determine the moles of C₈H₁₈ that could be combusted: 1 C₈H₁₈ / 12.5 O₂ = n(C₈H₁₈) / 4.60 × 10⁻³ mol O₂ n(C₈H₁₈) = (1/12.5) × 4.60 × 10⁻³ mol O₂ n(C₈H₁₈) = 3.68 × 10⁻⁴ mol C₈H₁₈ Finally, convert the moles of C₈H₁₈ to grams by multiplying by the molar mass of C₈H₁₈ (114.23 g/mol): mass (C₈H₁₈) = n(C₈H₁₈) × Molar Mass (C₈H₁₈) mass (C₈H₁₈) = 3.68 × 10⁻⁴ mol × 114.23 g/mol mass (C₈H₁₈) = 0.042 g The final answers: (a) There are 4.60 × 10⁻³ moles of O₂ present in the cylinder. (b) 0.042 grams of C₈H₁₈ could be combusted by this quantity of O₂, assuming complete combustion.

Key concepts

Mole Fraction

When we talk about the mole fraction, it's a way of expressing the concentration of a component in a mixture. It's a ratio that tells us how much of a particular substance is present compared to the total amount of all substances in the mixture. For example, in dry air, the mole fraction of oxygen (\(\mathrm{O_2}\)) is around 0.2095. This means that oxygen makes up about 20.95% of the air, in terms of moles.

This calculation becomes handy, especially when dealing with ideal gas law problems where you need to find out specific quantities of gases in a larger mixture. By knowing the mole fraction, you can simply multiply it by the total number of moles of the mixture to find out the moles of the component you are interested in.

Here, the ideal gas law is applied to find moles of air, then we use the mole fraction to find how many moles of oxygen are present. This ratio is crucial when considering reactions like combustion, where specific gases react in defined stoichiometric amounts.

Combustion Reaction

A combustion reaction is a chemical process in which a substance reacts with oxygen, releasing energy in the form of heat and light. This type of reaction is common in engines, like those found in vehicles. When fuel, such as octane (\(\mathrm{C_8H_{18}}\)), combusts, it reacts with oxygen to produce carbon dioxide and water as products.

The general form of a combustion reaction is:

• Fuel + Oxygen → Carbon Dioxide + Water

This simple representation hides the complex reactions, where balancing the number of atoms on both sides of the equation is essential. For example, the combustion reaction for octane is balanced as:\[\mathrm{C_8H_{18} + 12.5 \ O_2 \rightarrow 8 \ CO_2 + 9 \ H_2O}\]Balancing the chemical equation ensures that the conservation of mass is respected, meaning all atoms on the reactants side are present in equal numbers in the products. Understanding this concept helps predict how much fuel is needed for a certain amount of oxygen.

Stoichiometry

Stoichiometry is like the recipe of chemistry. It tells us how much of each reactant is required for a reaction and how much product will be formed. It's all about measuring and calculating the quantities of reactants and products in chemical reactions.

Stoichiometry uses the balanced chemical equations to determine the relationships between the different substances. For instance, from the balanced combustion reaction of octane, we see that 1 mole of octane reacts with 12.5 moles of oxygen. Such relationships are key to solving problems where you want to determine the amount of fuel combusted given a certain amount of oxygen.

By setting up ratios based on these stoichiometric coefficients, you can convert between moles of different substances. In our example, finding how much octane is burned by the available oxygen involves calculating a ratio and using molar masses to convert moles to grams. This method provides precise control over the reactants, essential in scenarios like fueling engines efficiently.