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 atm, how many moles of \(\mathrm{O}_{2}\) are present? (The mole fraction of \(\mathrm{O}_{2}\) in dry air is \(0.2095 . )(\mathbf{b})\) 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 approximately 0.00406 moles of O₂ present in the cylinder. (b) About 0.0371 grams of C₈H₁₈ could be combusted by this quantity of O₂, assuming complete combustion with the formation of CO₂ and H₂O.

Step-by-step solution

Convert the given cylinder volume to liters

The cylinder volume is given in cm³. We will convert it to L (liters) to match the units used for the Ideal Gas Law. \(Volume \ (L) = \frac{524 cm^3}{1000 cm^3/L} = 0.524 L\)

Calculate the total moles of air in the cylinder using the Ideal Gas Law

We will use the Ideal Gas Law (PV = nRT) to find the total number of moles of air (n) in the cylinder. \(n = \frac{PV}{RT}\) P = 0.980 atm (already in appropriate units) V = 0.524 L (found in Step 1) T = 74°C + 273.15 = 347.15 K (converted to Kelvin) R = 0.0821 L atm/(mol K) (Ideal Gas constant) \(n = \frac{(0.980 atm)(0.524 L)}{(0.0821 L atm/mol K)(347.15 K)} ≈ 0.0194 mol\)

Calculate the moles of O₂ present in the cylinder

Now we'll multiply the total moles of air by the mole fraction of O₂ (0.2095) to find the moles of O₂ present. Moles of O₂ = (0.0194 mol)(0.2095) ≈ 0.00406 mol

Determine how many moles of octane can be combusted by using the moles of O₂ and the balanced combustion reaction

The balanced combustion reaction is: C₈H₁₈ + 12.5 O₂ -> 8 CO₂ + 9 H₂O 1 mole of C₈H₁₈ requires 12.5 moles of O₂ for complete combustion. We have 0.00406 moles of O₂ present in the cylinder, so: Moles of C₈H₁₈ = (moles of O₂)/(12.5) Moles of C₈H₁₈ = (0.00406 mol) / (12.5) ≈ 0.000325 mol

Convert moles of C₈H₁₈ to grams

Lastly, we will convert the moles of C₈H₁₈ to grams using its molar mass (114.22 g/mol): Grams of C₈H₁₈ = (0.000325 mol)(114.22 g/mol) ≈ 0.0371 g Therefore, 0.0371 grams of C₈H₁₈ could be combusted by this quantity of O₂, assuming complete combustion with the formation of CO₂ and H₂O.

Key concepts

Combustion Reaction

A combustion reaction is a chemical reaction where a substance combines with oxygen and releases energy in the form of heat and light. It’s the process that powers engines and heats homes.
In the context of fuels like octane (a component of gasoline), combustion typically results in the formation of carbon dioxide (CO₂) and water (H₂O), along with the release of energy.
When dealing with chemical reactions, it’s crucial to balance the chemical equation to reflect the correct proportions of reactants and products. For instance, the combustion of octane is represented by the reaction: \[ \text{C}_8\text{H}_{18} + 12.5 \text{O}_2 \rightarrow 8 \text{CO}_2 + 9 \text{H}_2\text{O} \]This equation tells us that one mole of octane reacts with 12.5 moles of oxygen. During the process, energy is released, which is harnessed in automobile engines to move vehicles. Understanding this reaction is essential because it helps in determining how much fuel is needed and the amount of oxygen required for effective combustion.

Mole Concept

The mole is a fundamental unit in chemistry used to quantify amount of substance. It allows chemists to count atoms, molecules, and ions in a given sample with ease.
One mole is defined as exactly 6.02214076 × 10²³ particles (Avogadro's number). This large number bridges the micro-world of atoms to the scale of substances we can handle and measure.
The Ideal Gas Law often uses moles to express the amount of gas: \[ n = \frac{PV}{RT} \]Where:

• \(P\) is the pressure
• \(V\) is the volume
• \(n\) is the amount of substance in moles
• \(R\) is the ideal gas constant
• \(T\) is the temperature

Using this law, we can find how many moles of air are in the cylinder by given conditions. Understanding the mole concept allows us to interpret and predict the outcomes of chemical reactions and relate to the quantity of products formed, such as how much fuel can be burned given a certain amount of oxygen.

Stoichiometry

Stoichiometry is a section of chemistry that involves calculating relations between reactants and products in a chemical reaction. It is based on the law of conservation of mass where the mass of the reactants equals the mass of the products.
In the equation of the combustion of octane, stoichiometry helps us convert the moles of oxygen to moles of octane, and from there, to grams of octane.Through the balanced chemical equation, we know that \(12.5\) moles of \(\text{O}_2\) are needed for every mole of \(\text{C}_8\text{H}_{18}\).

• We found we had \(0.00406\) moles of \(\text{O}_2\).
• Therefore, the moles of \(\text{C}_8\text{H}_{18}\) that can be combusted is \(\frac{0.00406}{12.5} \approx 0.000325\) moles.
• These moles were then converted to grams by multiplying by the molar mass of \(\text{C}_8\text{H}_{18}\), which is \(114.22\) g/mol.

Stoichiometry is crucial for quantitative analysis in chemistry, helping to ensure reactions are efficient and that the correct proportions of substances are used.